When a blind beetle crawls over the surface of a curved branch, it doesn't notice that the track it has covered is indeed curved. I was lucky enough to notice what the beetle didn't notice.
Albert Einstein
In everyday life, if an object shows some degree of curvature (like the surface of a bowl, for example) we know about the curvature because the object is embedded in the three-dimensional Euclidean space of our everyday experience. ^(1){ }^{1} Mathematically, we can describe the geometry of the two-dimensional surface of the bowl using the three coordinates of the Euclidean space. It might therefore seem that a natural way to describe curved spacetime would be to embed it in a higher dimensional Euclidean space. However, there is absolutely no evidence that this is how Nature works and so, to avoid introducing unnecessary structure to the geometry of spacetime, we seek a way to describe curved space using only the coordinates available to a geometer confined to that space. Our fate, as geometers of a spacetime that we cannot step outside of is, therefore, a little like that of a beetle walking on the curved surface of the bowl. If the beetle is unable to escape the surface, then the coordinates it uses to measure lengths between points on the bowl will be two-dimensional. Despite this, we can use the notion of embedding to (i) work out the metric of a curved space in terms of those coordinates that a trapped geometer would use, and (ii) given a metric in terms of the coordinates within the surface, to visualize the space by embedding it in a Euclidean space. In order to embed a space, you need more than one more dimension in the Euclidean space than are used in the object's metric. ^(2){ }^{2}
In order to tackle embedding, we start by addressing a related problem. Say we already have an object in Euclidean space. We can then work out the metric on its surface in terms of the coordinates confined within that surface. We do this by relating the Euclidean coordinates to the internal, or object coordinates, and the expression for the metric we then obtain, in terms of the object's coordinates, is known as an induced metric.
Recall that the distance between points in Euclidean 3 -space R^(3)\mathbb{R}^{3} with coordinates X^(alpha)=(X,Y,Z)X^{\alpha}=(X, Y, Z) is given by
Chapter summary 585
Exercises ^(1){ }^{1} Our discussion in this Appendix follows Zee, which can be consulted for further details. ^(2){ }^{2} John Nash (1928-2015) is perhaps most famous for his Nobel Memorial Prize-winning work in Economics. His work in geometry includes the Nash embedding theorems, which demonstrate that any Riemannian manifold can be embedded in a Euclidean space. Silvia Nasar's biography of Nash, A Beautiful Mind, comes highly recommended, but should not be confused with the film of the same name, with which it bears little resemblance.
for NN-dimensional Euclidean space R^(N)\mathbb{R}^{N}. In what follows, we shall embed objects from DD-dimensional spaces into R^(N)\mathbb{R}^{N}. There are two methods to consider to find the induced metric.
Method II is suitable for when we have an equation for each of the Euclidean coordinates X^(alpha)X^{\alpha} in terms of the coordinates of the DD-dimensional object x^(mu)x^{\mu}. We write the coordinates X^(alpha)(x^(1)dotsx^(D))X^{\alpha}\left(x^{1} \ldots x^{D}\right), where alpha=1,dots,N\alpha=1, \ldots, N. We start by saying
where, in the final line, we've noted the general rule that ds^(2)=\mathrm{d} s^{2}=g_(mu nu)dx^(mu)dx^(nu)g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}. We deduce that the relationship between the metric in the object space g_(mu nu)g_{\mu \nu} is related to the embedding coordinates via
Method II is suitable for use when you can express the shape of the object you're trying to represent in Euclidean coordinates. Again we start with the metric line element in Euclidean space ds^(2)=dX^(2)+\mathrm{d} s^{2}=\mathrm{d} X^{2}+dY^(2)+dZ^(2)\mathrm{d} Y^{2}+\mathrm{d} Z^{2}. We then write an equation describing the object Z=f(X,Y)Z=f(X, Y) in terms of the embedding coordinates, which allows us to eliminate a coordinate ZZ. We then require an educated guess at some good objectspace coordinates x^(mu)x^{\mu} to reduce the metric to something convenient.
Example D. 1
(a) Consider a circle. If we confine ourselves to the surface of the circle it is a onedimensional space, so D=1D=1. The coordinate of particular points on the circle can be specified by giving an object coordinate theta\theta.
Start with method I. We can mount a circle in Euclidean ( N=2N=2 )-space (see Fig. D.1) by writing
{:(D.6)X=a cos thetaquad" and "quad Y=a sin theta:}\begin{equation*}
X=a \cos \theta \quad \text { and } \quad Y=a \sin \theta \tag{D.6}
\end{equation*}
where aa is the radius of the circle in the Euclidean space. We have therefore written a relationship between the circle's coordinate theta\theta and the embedding coordinates X^(alpha)X^{\alpha}. Differentiating, we have
{:(D.7)(del X)/(del theta)=-a sin thetaquad" and "quad(del Y)/(del theta)=a cos theta:}\begin{equation*}
\frac{\partial X}{\partial \theta}=-a \sin \theta \quad \text { and } \quad \frac{\partial Y}{\partial \theta}=a \cos \theta \tag{D.7}
\end{equation*}
This enables us to read out the metric component using eqn D. 5 as
(b) Let's now go up a dimension and embed the unit D=2D=2 sphere in R^(3)\mathbb{R}^{3} (see Fig. D.2). We need two object-space coordinates (x^(1),x^(2))=(theta,phi)\left(x^{1}, x^{2}\right)=(\theta, \phi). We know a parametrization in R^(3)\mathbb{R}^{3} using spherical coordinates with r=1r=1 we write (X,Y,Z)=(X, Y, Z)=(sin theta cos phi,sin theta sin phi,cos theta)(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta). We plug in to find the metric induced by the embedding
and find g_(phi theta)=0g_{\phi \theta}=0. We conclude that the D=2D=2 sphere is described by the induced line element
{:(D.16)ds^(2)=dtheta^(2)+sin^(2)thetadphi^(2).:}\begin{equation*}
\mathrm{d} s^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2} . \tag{D.16}
\end{equation*} Let's now try method II. The equation of the spherical surface in R^(3)\mathbb{R}^{3} is X^(2)+Y^(2)+X^{2}+Y^{2}+Z^(2)=1Z^{2}=1. Eliminating ZZ we have
Now to check for consistency with some well-chosen coordinates. The metric looks cylindrically symmetric, so we replace (X,Y)(X, Y) with cylindrical coordinates. Use X=X=r cos theta,Y=r sin theta,X^(2)+Y^(2)=r^(2)r \cos \theta, Y=r \sin \theta, X^{2}+Y^{2}=r^{2} and XdX+YdY=rdrX \mathrm{~d} X+Y \mathrm{~d} Y=r \mathrm{~d} r. We find
R^(3)\mathbb{R}^{3}. D. 2 The 2-sphere, embedded in R^(3)\mathbb{R}^{3}.
This looks different to the method-I answer, until we realize we can write r=sin thetar=\sin \theta, then dr^(2)=cos^(2)thetadtheta^(2)\mathrm{d} r^{2}=\cos ^{2} \theta \mathrm{~d} \theta^{2} and so ds^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} s^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2} as we had before.
Let's now turn to the problem of visualizing metrics. Often we are faced with the situation where we are presented with a line element and we want to know what the surface looks like when it's embedded in a higher dimensional Euclidean space. The next example shows an example of how to do this for the Einstein-Rosen bridge.
Fig. D. 3 The function Z(r)=Z(r)=+-[4r_(S)(r-r_(S))]^((1)/(2))\pm\left[4 r_{\mathrm{S}}\left(r-r_{\mathrm{S}}\right)\right]^{\frac{1}{2}}.
Fig. D. 4 The wormhole embedded in R^(3)\mathbb{R}^{3}. ^(3){ }^{3} These regions are called asymptotically flat: they are flat at large distances away from the throat.
Fig. D. 5 (a) The wormhole embedded in R^(3)\mathbb{R}^{3} showing a possible spacetime topology. (b) A topologically identical spacetime, showing the wormhole linking two distant points in spacetime. ^(4){ }^{4} Like the Alcubierre warp drive from Chapter 5, the creation of a wormhole requires negative energies.
Example D. 2
The Einstein-Rosen bridge (also known as a wormhole) is described by a D=2D=2 metric
We try a set of coordinates (X^(1),X^(2),X^(3))=(r cos theta,r sin theta,Z(r))\left(X^{1}, X^{2}, X^{3}\right)=(r \cos \theta, r \sin \theta, Z(r)), which solve the second equation, as can be easily seen. The first equation then becomes
which is solved if we set Z(r)^(2)=4r_(S)(r-r_(S))Z(r)^{2}=4 r_{\mathrm{S}}\left(r-r_{\mathrm{S}}\right). The function Z(r)Z(r) is graphed in Fig. D.3. If it is rotated about the zz-axis it generates the spacetime shown in Fig. D.4. The wormhole metric, embedded in R^(3)\mathbb{R}^{3} appears to show two regions of flat ^(3){ }^{3} spacetime joined by a throat. (The narrow hole down the middle explains the name wormhole, of course.) The metric tells us about the detailed geometry of spacetime, but tells us nothing about the topology of the spacetime, that is, the large scale shape and us nothing about the topology of the spacetime, that is, the large scale shape and
connectivity of the space. As a result, the wormhole might link two completely connectivity of the space. As a result, the wormhole might link two completely
different parts of a spacetime. One possibility is shown in Fig. D.5(a), where the wormhole links two distant points in a spacetime. A topologically identical diagram is shown in Fig. D.5(b), revealing that two distant patches of flat spacetime are linked by the wormhole. If the distance between the two regions via the flat space is greater than the distance through the throat, then we potentially have a shortcut between very distant points in spacetime. It has also been suggested that the wormhole might joint two disconnected parts of spacetime, allowing an explorer to visit a different Universe entirely. ^(4){ }^{4}
There are some surfaces that cannot be embedded in Euclidean space. The most important example for us is hyperbolic space.
We can try some coordinates for XX and YY that retain the cylindrical symmetry of this space, and determine the ZZ coordinate. We try
{:(D.24)X=sinh chi cos phi","quad Y=sinh chi sin phi:}\begin{equation*}
X=\sinh \chi \cos \phi, \quad Y=\sinh \chi \sin \phi \tag{D.24}
\end{equation*}
Plugging in to the second equation, we see that del Z//del phi=0\partial Z / \partial \phi=0, so ZZ is indeed cylindrically symmetric. The first equation tells us that
This equation has no real solutions. The embedding has failed.
This embedding is indeed impossible. There are two options, the first is simply to embed part of the surface. ^(5){ }^{5} Perhaps a more satisfactory solution is not to mount the space in Euclidean space, but rather in some other space. For hyperbolic spaces, the natural higher dimensional space is known as pseudo-Euclidean space which has a metric
(a) Consider a surface X^(2)+Y^(2)-W^(2)=-1X^{2}+Y^{2}-W^{2}=-1, known as a 2 -sheet hyperboloid. This can be embedded in pseudo-Euclidean space. ^(6){ }^{6} Use the coordinates
{:(D.27)X=sinh chi cos theta","quad Y=sinh chi sin theta","quad W=cosh chi:}\begin{equation*}
X=\sinh \chi \cos \theta, \quad Y=\sinh \chi \sin \theta, \quad W=\cosh \chi \tag{D.27}
\end{equation*}
which are consistent with the equation for the surface. Differentiating, we find that
This is the metric we considered before. We conclude that the embedding is possible in the pseudo-Euclidean metric. The surface defined by these coordinates is shown in the Fig. D. 6.
(b) Now try the surface X^(2)+Y^(2)-W^(2)=1X^{2}+Y^{2}-W^{2}=1, known as the one-sheet hyperboloid. This one is solved for the coordinate choice
{:(D.29)X=cosh chi cos theta","quad Y=cosh chi sin theta","quad W=sinh chi:}\begin{equation*}
X=\cosh \chi \cos \theta, \quad Y=\cosh \chi \sin \theta, \quad W=\sinh \chi \tag{D.29}
\end{equation*}
which is simply the metric for the 2 -sphere we had before. In fact, we could have generated this metric directly with the choice of coordinates
{:(D.32)X=r cos theta","quad Y=r sin theta","quad W=(r^(2)-1)^((1)/(2)):}\begin{equation*}
X=r \cos \theta, \quad Y=r \sin \theta, \quad W=\left(r^{2}-1\right)^{\frac{1}{2}} \tag{D.32}
\end{equation*}
The spherical line element that represents the one-sheet hyperboloid should not be misinterpreted to mean that the shape of the space is a sphere. We have eliminated WW leaving a set of concentric circles behind, with distances encoded in the difference between the circles. (Put another way, we have sliced the spacetime leaving circular cross sections.) If we set r=sin thetar=\sin \theta as we did for the 2 -sphere then we would not be able to solve the equation that defines the surface: r^(2)-W^(2)=1r^{2}-W^{2}=1. Setting r=cosh chir=\cosh \chi does not cause this problem. See Chapter 18 for more details on this space which is very useful in cosmology. method I to this case.
Fig. D. 6 The 2-sheet hyperboloid embedded in pseudo-Euclidean space.
Chapter summary
Embedding allows us to visualize a metric space by mounting it in a higher dimensional Euclidean space.
Exercises
(D.1) A parabolic 2-surface in Euclidean 3-space (see (D.3) The D=2D=2 torus may be embedded in R^(3)\mathbb{R}^{3} (Fig. D.8) Fig. D.7) is described by
We saw in Example D.3 how embedding fails for a hyperbolic surface. However, we can embed part of this surface. Consider an alternative metric for a hyperbolic surface
This equation does enjoy real solutions for sinh chi <\sinh \chi< 1 , corresponding to the range 0 <= chi <= sinh^(-1)10 \leq \chi \leq \sinh ^{-1} 1. So a partial embedding of this surface is possible. This wasn't the hyperbolic line element we started with, but like that line element, it does have a constant negative curvature. Minding's theorem [named after Ferdinand Minding (1806-1885)J says that all constant-curvature surfaces have the same local geometry, and so the geometry of this line element is representative of the hyperbolic spacetime.
(D.5) Consider the Rindler coordinate system (x^(1),x^(2))=\left(x^{1}, x^{2}\right)=(t,x)(t, x), embedded in a two-dimensional Minkowski space (T,X)(T, X) via
{:(D.38)T=x sinh t","quad X=x cosh t:}\begin{equation*}
T=x \sinh t, \quad X=x \cosh t \tag{D.38}
\end{equation*}
Compute the components of the induced metric. Hint: Since we want to work in Minkowski space, rather than Euclidean space, we must use
(0.1) Equating kinetic energy (1)/(2)mv_("esc ")^(2)\frac{1}{2} m v_{\text {esc }}^{2} to the gravitational potential energy GMm//rG M m / r yields v_("esc ")=sqrt(2GM//r)v_{\text {esc }}=\sqrt{2 G M / r} and since Phi=-GMm//r\Phi=-G M m / r then we also have that v_("esc ")=sqrt(2|Phi|)v_{\text {esc }}=\sqrt{2|\Phi|}. Rearranging for v_("esc ")=cv_{\text {esc }}=c gives the Schwarzschild radius r=2GM//c^(2)r=2 G M / c^{2}.
(0.2) (i) Earth: 9.8ms^(-2);11.2kms^(-1)(~~0.00004 c)9.8 \mathrm{~m} \mathrm{~s}^{-2} ; 11.2 \mathrm{~km} \mathrm{~s}^{-1}(\approx 0.00004 c); (ii) Sun: 274ms^(-2);6.2 xx10^(5)ms^(-1)(~~0.002 c)274 \mathrm{~m} \mathrm{~s}^{-2} ; 6.2 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}(\approx 0.002 c); (iii) neutron star: 1.9 xx10^(12)ms^(-2);1.9 xx10^(8)ms^(-1)(~~0.64 c)1.9 \times 10^{12} \mathrm{~m} \mathrm{~s}^{-2} ; 1.9 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}(\approx 0.64 c).
(0.3) The inverse square law of force in Newtonian gravity implies that |Delta F|//|F|=2|Delta r|//|r||\Delta F| /|F|=2|\Delta r| /|r| and so with Delta r=1.8m\Delta r=1.8 \mathrm{~m} we have that for the Earth the effect of the tidal force is |Delta F|//|F|=2xx1.8m//R_(o+)|\Delta F| /|F|=2 \times 1.8 \mathrm{~m} / R_{\oplus} which is around 0.6 ppm , i.e. the tidal force is less than a millionth of gravitational force, which is why you don't notice it. The Schwarzschild radius for a 3M_(o.)3 M_{\odot} black hole is about 9 km , so |Delta F|//|F|~~0.0004|\Delta F| /|F| \approx 0.0004 (though the tidal forces are enormous since the gravitational field is more than eleven orders of magnitude bigger than it is on Earth). The Schwarzschild radius for a than it is on Earth). The Schwarzschild radius for a 10^(6)M_(o.)10^{6} M_{\odot} black hole is about three million km , and hence 10^(6)M_(o.)10^{6} M_{\odot} black hole is about three million km, and hence |Delta F|//|F|~~4xx10^(-10)|\Delta F| /|F| \approx 4 \times 10^{-10}. The tidal forces are smaller in |Delta F|//|F|~~4xx10^(-10)|\Delta F| /|F| \approx 4 \times 10^{-10}. The tidal forces are smaller in hole is only about six orders of magnitude greater than on Earth.
(At the Schwarzschild radius, since r prop Mr \propto M then |Delta F|prop|\Delta F| \proptoF//rF / r and since F prop M//r^(2)F \propto M / r^{2} then |Delta F|propM^(-2)|\Delta F| \propto M^{-2}, so that tidal forces at the Schwarzschild radius are larger for small black holes than for supermassive black holes.)
(1.2) gamma=(1-0.995^(2))^(-1//2)~~10\gamma=\left(1-0.995^{2}\right)^{-1 / 2} \approx 10. (a) 0.995 c xx2.2 mus=0.66km0.995 c \times 2.2 \mu \mathrm{~s}=0.66 \mathrm{~km}. (b) Multiply 0.66 km by gamma:6.6km\gamma: 6.6 \mathrm{~km}.
(1.3) The interval is given by Deltas^(2)=-c^(2)Deltat^(2)+Deltax^(2)\Delta s^{2}=-c^{2} \Delta t^{2}+\Delta x^{2}. The time interval Delta t\Delta t is given by the difference between the time of events qq and pp as measured by us using coordinates (t,x)(t, x). Take the time coordinate of event rr to be the origin. This means the event pp takes place at a time t_(p)=tau_(2)t_{p}=\tau_{2}. Since the light takes a time tau_(2)+tau_(1)\tau_{2}+\tau_{1} to reach qq and return, we reason that the event qq takes place at a time t_(q)=(tau_(2)+tau_(1))//2t_{q}=\left(\tau_{2}+\tau_{1}\right) / 2. As a result we have that
In order to work out the interval in position Delta x\Delta x, we note that the distance travelled from our world line to
the event qq is measured by the light pulse as Delta x=\Delta x=c(tau_(1)+tau_(2))//2c\left(\tau_{1}+\tau_{2}\right) / 2. Plugging into the expression for the interval Deltas^(2)\Delta s^{2}, we obtain
(1.4) (a) 0.140 , (b) 0.417 , (c) 0.866 , (d) 0.996 , (e) 0.99995 .
(2.1) This is a straightforward application of eqn 2.38 .
(2.2) Write u=gamma( vec(u))(1, vec(u))\boldsymbol{u}=\gamma(\vec{u})(1, \vec{u}) and v=gamma( vec(v))(1, vec(v))\boldsymbol{v}=\gamma(\vec{v})(1, \vec{v}), so that u*v=gamma( vec(u))gamma( vec(v))(-1+ vec(u)* vec(v))\boldsymbol{u} \cdot \boldsymbol{v}=\gamma(\vec{u}) \gamma(\vec{v})(-1+\vec{u} \cdot \vec{v}) and this is equal to -gamma-\gamma, yielding the result. In the non-relativistic limit, gamma=(1-v_(rel)^(2))^(-1//2)=1+v_(rel)^(2)//2+cdots\gamma=\left(1-v_{\mathrm{rel}}^{2}\right)^{-1 / 2}=1+v_{\mathrm{rel}}^{2} / 2+\cdots and hence
and the two results follow after taking the square root. This is just the Galilean relative velocity, and the two results are found because we have not specified whether the relative velocity is uu relative to vv or the other way around.
(2.3) The first answers follow by simple matrix multiplication using eta_(mu nu)\eta_{\mu \nu} given in eqn 2.15. Some of the later answers are given in the main text of the chapter.
(2.4) The proton has gamma=10^(19)//10^(9)=10^(10)\gamma=10^{19} / 10^{9}=10^{10} and so is travelling very close to the speed of light. It travels 10^(5)10^{5} light-years in 10^(5)10^{5} years. In its rest frame, the galaxy is contracted by a factor gamma\gamma and so the journey is only 10^(-5)10^{-5} light-years long, taking 10^(-5)10^{-5} years, or about 5 minutes.
(2.5) The Lorentz transformation gives p^(mu)=(gamma m,gamma mv,0,0)p^{\mu}=(\gamma m, \gamma m v, 0,0) and so one can just read off E=gamma mE=\gamma m (i.e. E=gamma mc^(2)E=\gamma m c^{2} if you put the factors of cc back in) and p^(x)=gamma mvp^{x}=\gamma m v. Both sets of components give p*p=-(p^(0))^(2)+(p^(1))^(2)+(p^(2))^(2)+\boldsymbol{p} \cdot \boldsymbol{p}=-\left(p^{0}\right)^{2}+\left(p^{1}\right)^{2}+\left(p^{2}\right)^{2}+(p^(3))^(2)=-m^(2)\left(p^{3}\right)^{2}=-m^{2}.
(2.6) Assume that we can have e+gamma rarre\mathrm{e}+\gamma \rightarrow \mathrm{e}, so that
If we square both sides, we get p_(e)*p_(e)+2p_(e)*p_(gamma)+p_(gamma)*p_(gamma)=p_(e)^(')*p_(e)^(')\boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\mathrm{e}}+2 \boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\gamma}+\boldsymbol{p}_{\gamma} \cdot \boldsymbol{p}_{\gamma}=\boldsymbol{p}_{\mathrm{e}}^{\prime} \cdot \boldsymbol{p}_{\mathrm{e}}^{\prime},
but p_(e)*p_(e)=p_(e)^(')*p_(e)^(')=-m_(e)^(2)\boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\mathrm{e}}=\boldsymbol{p}_{\mathrm{e}}^{\prime} \cdot \boldsymbol{p}_{\mathrm{e}}^{\prime}=-m_{\mathrm{e}}^{2} and p_(gamma)*p_(gamma)=0\boldsymbol{p}_{\gamma} \cdot \boldsymbol{p}_{\gamma}=0, so we are left with 2p_(e)*p_(gamma)=02 \boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\gamma}=0. We can evaluate this in the rest frame of the electron where p_(e)=(m_(e), vec(0))\boldsymbol{p}_{\mathrm{e}}=\left(m_{\mathrm{e}}, \overrightarrow{0}\right) and in general the photon has p_(gamma)=(E, vec(p))\boldsymbol{p}_{\gamma}=(E, \vec{p}) with E=| vec(p)|E=|\vec{p}|. So we are left with Em_(e)=0E m_{\mathrm{e}}=0, which means that E=0E=0, i.e. only holds for (non-existent) zero-energy photons.
(2.7) Here the coordinate qq of interest is q=phiq=\phi, the amplitude of the string's displacement. However, we're not interested in phi\phi as a function of time, but rather of length xx along the string. This is no problem, we simply proceed with phi^(˙)=del phi//del x\dot{\phi}=\partial \phi / \partial x. Instead of the action, the functional we want to minimize is the energy E[phi]E[\phi]. There are two contributions to the energy, the strain propdphi//dx\propto \mathrm{d} \phi / \mathrm{d} x tells us it costs energy for the string to bend, the gravitational potential energy V=-rho g phi(x)V=-\rho g \phi(x) shows that the energy is reduced the large xx. The energy functional is
{:(E.7)E(phi)=intdx[(T)/(2)(((d)phi(x))/(dx))^(2)-rho g phi(x)]:}\begin{equation*}
E(\phi)=\int \mathrm{d} x\left[\frac{T}{2}\left(\frac{\mathrm{~d} \phi(x)}{\mathrm{d} x}\right)^{2}-\rho g \phi(x)\right] \tag{E.7}
\end{equation*}
Notice that this is in the same form as the mechanical action, with the first term playing the role of the kinetic energy. Feeding this through the Euler Lagrange equations, we find
(4.1) Since u\boldsymbol{u} and v\boldsymbol{v} transform as 4 -vectors, the tensor transformation law gives Lambda^(alpha^('))_(alpha)Lambda^(beta^('))_(beta)W^(alpha beta)=Lambda^(alpha^('))_(alpha)^(beta^('))_(beta)u^(alpha)v^(beta)=\Lambda^{\alpha^{\prime}}{ }_{\alpha} \Lambda^{\beta^{\prime}}{ }_{\beta} W^{\alpha \beta}=\Lambda^{\alpha^{\prime}}{ }_{\alpha}{ }^{\beta^{\prime}}{ }_{\beta} u^{\alpha} v^{\beta}=u^(alpha^('))v^(beta^('))=W^(alpha^(')beta^('))u^{\alpha^{\prime}} v^{\beta^{\prime}}=W^{\alpha^{\prime} \beta^{\prime}} as required.
(4.2) Lambda^(alpha^('))_(alpha)Lambda^(beta)_(beta^('))delta^(alpha)_(beta)=Lambda^(alpha^('))_(beta)Lambda^(beta)_(beta^('))=delta^(alpha^('))_(^('))\Lambda^{\alpha^{\prime}}{ }_{\alpha} \Lambda^{\beta}{ }_{\beta^{\prime}} \delta^{\alpha}{ }_{\beta}=\Lambda^{\alpha^{\prime}}{ }_{\beta} \Lambda^{\beta}{ }_{\beta^{\prime}}=\delta^{\alpha^{\prime}}{ }_{{ }^{\prime}} as required for it to transform properly. We can use the metric tensor to raise and lower the indices in delta^(alpha beta)\delta^{\alpha \beta} and delta_(alpha beta)\delta_{\alpha \beta} as appropriate to turn them into delta^(alpha)_(beta)\delta^{\alpha}{ }_{\beta} (which we've just shown works as a tensor)
Thus, if you try and write delta_(alpha beta)\delta_{\alpha \beta} and delta^(alpha beta)\delta^{\alpha \beta} as tensors, they end up being equal to the metric tensor and lose their Kronecker-delta property. Thus, it is best to keep delta_(alpha beta)\delta_{\alpha \beta} and delta^(alpha beta)\delta^{\alpha \beta} as non-tensor objects which are shorthands for quantities equalling one when alpha=beta\alpha=\beta and zero otherquantities equalling one when alpha=beta\alpha=\beta if you want the Kronecker delta to behave as a
wise. wise. If you want the Kronecker delta to be
tensor, you have to use the mixed form delta^(alpha)_(beta)\delta^{\alpha}{ }_{\beta}.
(4.3) S\boldsymbol{S} is a tensor and so its components transforms properly
{:[x=r sin theta cos phi","],[y=r sin theta sin phi","],[(E.22)z=r cos theta]:}\begin{align*}
& x=r \sin \theta \cos \phi, \\
& y=r \sin \theta \sin \phi, \\
& z=r \cos \theta \tag{E.22}
\end{align*}
Using the transformation law for basis vectors e_(mu)=e_{\mu}=(delx^(nu)//delx^(mu))e_(nu)\left(\partial x^{\nu} / \partial x^{\mu}\right) \boldsymbol{e}_{\nu}, we find that the basis vectors are
{:[e_(r)=sin theta cos phie_(x)+sin theta sin phie_(y)+cos thetae_(z)],[e_(theta)=cos theta cos phie_(x)+cos theta sin phie_(y)-sin thetae_(z)],[e_(phi)=-r sin theta sin phie_(x)+r sin theta cos phie_(y)]:}\begin{aligned}
& \boldsymbol{e}_{r}=\sin \theta \cos \phi \boldsymbol{e}_{x}+\sin \theta \sin \phi \boldsymbol{e}_{y}+\cos \theta \boldsymbol{e}_{z} \\
& \boldsymbol{e}_{\theta}=\cos \theta \cos \phi \boldsymbol{e}_{x}+\cos \theta \sin \phi \boldsymbol{e}_{y}-\sin \theta \boldsymbol{e}_{z} \\
& \boldsymbol{e}_{\phi}=-r \sin \theta \sin \phi \boldsymbol{e}_{x}+r \sin \theta \cos \phi \boldsymbol{e}_{y}
\end{aligned}
(c) Notice that the components for the vector e_(r)\boldsymbol{e}_{r} and 1 -form omega^(r)\boldsymbol{\omega}^{r} are the same; those for theta\theta related by a factor rr and those for phi\phi by a factor r^(2)sin^(2)thetar^{2} \sin ^{2} \theta. The orthogonality condition (:omega^(mu),e_(nu):)=delta^(mu)_(nu)\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\nu}\right\rangle=\delta^{\mu}{ }_{\nu} is then guaranteed by the orthogonality of the basis vectors e_(nu)\boldsymbol{e}_{\nu} (which can be checked).
(4.5) (a) Using e_(u)=(delx^(mu)//del u)e_(mu)\boldsymbol{e}_{u}=\left(\partial x^{\mu} / \partial u\right) \boldsymbol{e}_{\mu}, and so on, we find
{:[e_(u)=e_(x)+e_(y)-2ve_(z)","],[e_(v)=-e_(x)+e_(y)-2ue_(z)","],[(E.24)e_(w)=e_(z)]:}\begin{align*}
\boldsymbol{e}_{u} & =\boldsymbol{e}_{x}+\boldsymbol{e}_{y}-2 v \boldsymbol{e}_{z}, \\
\boldsymbol{e}_{v} & =-\boldsymbol{e}_{x}+\boldsymbol{e}_{y}-2 u \boldsymbol{e}_{z}, \\
\boldsymbol{e}_{w} & =\boldsymbol{e}_{z} \tag{E.24}
\end{align*}
(b) Using omega^(u)=(del u//delx^(mu))omega^(mu)\boldsymbol{\omega}^{u}=\left(\partial u / \partial x^{\mu}\right) \boldsymbol{\omega}^{\mu} and so on, we have
Setting k^(x)=| vec(k)|cos alpha=omega cos alphak^{x}=|\vec{k}| \cos \alpha=\omega \cos \alpha for a light source whose direction of emission makes an angle alpha\alpha with the xx direction, we have
{:(E.28)omega^(')=gamma omega(1-beta cos alpha):}\begin{equation*}
\omega^{\prime}=\gamma \omega(1-\beta \cos \alpha) \tag{E.28}
\end{equation*}
If the source is in the primed frame where it is measured to have a frequency omega_(0)\omega_{0}, we have
{:(E.29)(omega)/(omega_(0))=((1-beta^(2))^((1)/(2)))/(1-beta cos alpha):}\begin{equation*}
\frac{\omega}{\omega_{0}}=\frac{\left(1-\beta^{2}\right)^{\frac{1}{2}}}{1-\beta \cos \alpha} \tag{E.29}
\end{equation*}
(5.3) (b) Dropping primes, setting ds^(2)=0\mathrm{d} s^{2}=0 and dividing through by dt^(2)\mathrm{d} t^{2} we find, in the x-tx-t plane, that
{:(E.30)1-v^(2)=((dx)/((d)t))^(2)-2v((d)x)/((d)t):}\begin{equation*}
1-v^{2}=\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}-2 v \frac{\mathrm{~d} x}{\mathrm{~d} t} \tag{E.30}
\end{equation*}
Solving this quadratic equation, we obtain an equation for the light cones of
which tells us that the light cones make an angle tan^(-1)(v+-1)\tan ^{-1}(v \pm 1) to the tt axis.
(c) We obtain the usual expression for time dilation in
flat spacetime dtau=dt(1-v^(2))^((1)/(2))\mathrm{d} \tau=\mathrm{d} t\left(1-v^{2}\right)^{\frac{1}{2}}.
(d) We find
(5.5) (a) dl=dx\mathrm{d} l=\mathrm{d} x.
(b) dtau=xdt\mathrm{d} \tau=x \mathrm{~d} t, where xx is the position of the event.
(6.1) The time dilation factor is sqrt(1-2GM//rc^(2))\sqrt{1-2 G M / r c^{2}} and the numerical values come out to be (a) 1-1.1 xx10^(-8)1-1.1 \times 10^{-8} (b) 1-2.1 xx10^(-6)1-2.1 \times 10^{-6} (c) 0.84 . All three values are a bit less than 1, but for (c) it's quite a bit less. Note that for the time dilation factor on the surface of the Earth (comparing it to zero gravitational field) you have to include the effect of the Sun's gravitational field at the Earth's surface, so that the answer is obtained by computing sqrt(1-2G//c^(2)(M_(o+)//R_(o+)+M_(o.)//R))~~G//c^(2)(M_(o+)//R_(o+)+:}\sqrt{1-2 G / c^{2}\left(M_{\oplus} / R_{\oplus}+M_{\odot} / R\right)} \approx G / c^{2}\left(M_{\oplus} / R_{\oplus}+\right.M_(o.)//RM_{\odot} / R ), where RR is an astronomical unit.
(6.2) Differentiating the sqrt(1-2GM_(o+)//rc^(2))\sqrt{1-2 G M_{\oplus} / r c^{2}} factor gives -GM_(o+)//r^(2)c^(2)=-g//c^(2)=-1.1 xx10^(-16)m^(-1)=-1.1 xx-G M_{\oplus} / r^{2} c^{2}=-g / c^{2}=-1.1 \times 10^{-16} \mathrm{~m}^{-1}=-1.1 \times10^(-19)mm^(-1)10^{-19} \mathrm{~mm}^{-1}, which is consistent with the value given. (Note that, in comparison with the previous exercise, the effect of the Sun is not needed. Although the Sun's gravitational field leads to a substantial time dilation on the Earth's surface, its gradient is much smaller and can be neglected.)
(6.4) (a) Delta tau=(1-(2M)/(r))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} t.
(b) Delta tau=(1-(2M)/(r_(2)))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r_{2}}\right)^{\frac{1}{2}} \mathrm{~d} t, where dt\mathrm{d} t is the coordinate time interval between wavefront events.
(c) Since the observers are parallel, the interval between the wavefronts being received is the same as the interval between which they are emitted. As a result, the observer at r_(2)r_{2} measures an interval Delta tau=(1-(2M)/(r_(1)))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r_{1}}\right)^{\frac{1}{2}} \mathrm{~d} t.
(6.6) (a) Figure 6.6 shows the set up. The time on A's world line will be
The two intervals are the two solutions in eqn 6.15, whose sum gives the quoted integrand.
(c) The optical path length is 2pi r+2pi Omegar^(2)2 \pi r+2 \pi \Omega r^{2}. For a counter-propagating beam, the optical path is 2pi r-2 \pi r-2pi Omegar^(2)2 \pi \Omega r^{2}. The fringe shift is the difference in optical path, divided by the wavelength of the light.
7.1) (i) grad_(u)v-grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}.
(ii) grad_(v) tilde(w)(u)\boldsymbol{\nabla}_{\boldsymbol{v}} \tilde{\boldsymbol{w}}(\boldsymbol{u}).
(iii) In slot-machine notation, we could rewrite the expression (grad_(u)T)(v,w)\left(\nabla_{u} \boldsymbol{T}\right)(\boldsymbol{v}, \boldsymbol{w}), becoming u^(mu)T_(alpha beta;mu)v^(alpha)w^(beta)u^{\mu} T_{\alpha \beta ; \mu} v^{\alpha} w^{\beta}.
(7.2) (a) The components of the tangent are given by u^(mu)=u^{\mu}=(dx//ds,dy//ds)(\mathrm{d} x / \mathrm{d} s, \mathrm{~d} y / \mathrm{d} s) and so u^(mu)=(-sin s,cos s)u^{\mu}=(-\sin s, \cos s). This vector has unit length.
(b) The curve is not a geodesic, so we do not expect the components of the gradient to vanish. Differentiating, we find (Du//ds)^(mu)=(-cos s,-sin s)(\mathrm{D} \boldsymbol{u} / \mathrm{d} s)^{\mu}=(-\cos s,-\sin s), which gives zero
when dotted with u\boldsymbol{u}
(c) The tangent vector to the reparametrized curve has components u^(mu)=(-t//sqrt(1-t^(2)),1)u^{\mu}=\left(-t / \sqrt{1-t^{2}}, 1\right), from which we find u*u=(1-t^(2))^(-1)\boldsymbol{u} \cdot \boldsymbol{u}=\left(1-t^{2}\right)^{-1}. The gradient of this vector has components (Du//dt)^(mu)=(-(1-t^(2))^(-(3)/(2)),0)(\mathrm{D} \boldsymbol{u} / \mathrm{d} t)^{\mu}=\left(-\left(1-t^{2}\right)^{-\frac{3}{2}}, 0\right) and so u*Du//dt=t//(1-t^(2))^(2)\boldsymbol{u} \cdot \mathrm{D} \boldsymbol{u} / \mathrm{d} t=t /\left(1-t^{2}\right)^{2}
The fact that the tangent vector does not have a constant magnitude makes this parametrization less convenient than the (affine) length parametrization that uses ss.
(7.3) (a) We find
{:(E.36)(v^(r),v^(theta))=(Cr sin theta cos theta,Ccos^(2)theta):}\begin{equation*}
\left(v^{r}, v^{\theta}\right)=\left(C r \sin \theta \cos \theta, C \cos ^{2} \theta\right) \tag{E.36}
\end{equation*}
(c) The derivatives are
and
{:[(E.37)grad_(r)v=((C sin theta cos theta)/((C)/(r)*cos^(2)theta))","],[(E.38)grad_(theta)v=((-Crsin^(2)theta)/(-C cos theta sin theta)).]:}\begin{gather*}
\boldsymbol{\nabla}_{r} \boldsymbol{v}=\binom{C \sin \theta \cos \theta}{\frac{C}{r} \cdot \cos ^{2} \theta}, \tag{E.37}\\
\boldsymbol{\nabla}_{\theta} \boldsymbol{v}=\binom{-C r \sin ^{2} \theta}{-C \cos \theta \sin \theta} . \tag{E.38}
\end{gather*}
However, it's simplest to work in the other direction and compute (grad_(nu^('))v)^(mu^('))=Lambda^(x)_(nu^(')),Lambda^(mu^('))_(y)(grad_(x)v)^(y)\left(\boldsymbol{\nabla}_{\nu^{\prime}} \boldsymbol{v}\right)^{\mu^{\prime}}=\Lambda^{x}{ }_{\nu^{\prime}}, \Lambda^{\mu^{\prime}}{ }_{y}\left(\boldsymbol{\nabla}_{x} \boldsymbol{v}\right)^{y} for mu^('),nu^(')=r,theta\mu^{\prime}, \nu^{\prime}=r, \theta and check against the results in (c).
(8.1) (a) The E-L procedure gives
then eqn 8.48 becomes r=a//cos(theta-theta_(0))r=a / \cos \left(\theta-\theta_{0}\right), which shows that this is, indeed, a description of a straight line.
(8.3) First evaluate velocity
=(dx^(sigma))/(dtau)*(del^(2)xi^(mu))/(delx^(nu)delx^(sigma))(dx^(nu))/(dtau)+(dx^(sigma))/(dtau)*(delxi^(mu))/(delx^(nu))(del)/(delx^(sigma))(dx^(nu))/(dtau)=\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial^{2} \xi^{\mu}}{\partial x^{\nu} \partial x^{\sigma}} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}+\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial \xi^{\mu}}{\partial x^{\nu}} \frac{\partial}{\partial x^{\sigma}} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} =(dx^(sigma))/(dtau)*(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma))*(dx^(lambda))/(dtau)+(delxi^(mu))/(delx^(nu))(d^(2)x^(nu))/(dtau^(2))=0,quad=\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}} \cdot \frac{\mathrm{d} x^{\lambda}}{\mathrm{d} \tau}+\frac{\partial \xi^{\mu}}{\partial x^{\nu}} \frac{\mathrm{d}^{2} x^{\nu}}{\mathrm{d} \tau^{2}}=0, \quad (E.52)
where, in the first term of the last line, we sum over lambda\lambda instead of nu\nu to avoid confusion in the next stage. Tidying, we have
(b) Start with u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1, which implies d//dtau(u*u)=0\mathrm{d} / \mathrm{d} \tau(\boldsymbol{u} \cdot \boldsymbol{u})=0, from which we say
The connection coefficients are Gamma^(t)_(xt)=1//x\Gamma^{t}{ }_{x t}=1 / x and Gamma^(x)_(tt)=x\Gamma^{x}{ }_{t t}=x. (b) Using the chain rule we write
The length of the so-called maximal geodesic, that starts at t=0t=0 and ends at t=pit=\pi, is infinite.
(9.9) Call y=(-g_(mu nu)x^(˙)^(mu)x^(˙)^(nu))^((1)/(2))y=\left(-g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}\right)^{\frac{1}{2}}. The Euler-Lagrange equation gives
For length parametrization we have dy//dlambda=0\mathrm{d} y / \mathrm{d} \lambda=0, wiping out the first term. We also have that dF//dy!=0\mathrm{d} F / \mathrm{d} y \neq 0, as the function is monotonic. This then results in the same Euler-Lagrange equation as obtained from varying yy.
(9.11) We find connection coefficients
(10.6) (a) Using eta_( hat(mu)nu)=e_( hat(mu))*e_( hat(nu))\eta_{\hat{\mu} \nu}=\boldsymbol{e}_{\hat{\mu}} \cdot \boldsymbol{e}_{\hat{\nu}} we find
Differentiating with respect to lambda\lambda, the
Differentiating with respect to lambda\lambda, the velocity has components u^(mu)=(dr//dlambda,dphi//dlambda)u^{\mu}=(\mathrm{d} r / \mathrm{d} \lambda, \mathrm{d} \phi / \mathrm{d} \lambda) given by
and the other components vanish. Contracting against the components of the velocity then gives a vanishing acceleration.
(10.1) (a) We obtain
A^( hat(mu))=(A^(t),aA^(chi),a sinh chiA^(theta),a sinh chi sin thetaA^(phi))A^{\hat{\mu}}=\left(A^{t}, a A^{\chi}, a \sinh \chi A^{\theta}, a \sinh \chi \sin \theta A^{\phi}\right)
(10.2) (a) We find Gamma_( hat(theta) hat(theta))^( hat(theta))=-1//r,Gamma_( hat(theta) hat(r))^( hat(theta))=1//r\Gamma_{\hat{\theta} \hat{\theta}}^{\hat{\theta}}=-1 / r, \Gamma_{\hat{\theta} \hat{r}}^{\hat{\theta}}=1 / r and Gamma_( hat(theta) hat(theta))^( hat(theta))=0\Gamma_{\hat{\theta} \hat{\theta}}^{\hat{\theta}}=0. (b) From Example 3.3, we have that [e_( hat(r)),e_( hat(theta))]=^( hat(r) hat(theta))-e_( hat(theta))//r\left[\boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}\right] \stackrel{\hat{r} \hat{\theta}}{=}-\boldsymbol{e}_{\hat{\theta}} / r, and so (:omega^( hat(theta)),[e_( hat(r)),e_( hat(theta))]:)=-1//r\left\langle\boldsymbol{\omega}^{\hat{\theta}},\left[\boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}\right]\right\rangle=-1 / r, showing that the rule is obeyed for Gamma_( hat(hat(r)) hat(theta))-Gamma^( hat(theta))_( hat(theta) hat(r))\Gamma_{\hat{\hat{r}} \hat{\theta}}-\Gamma^{\hat{\theta}}{ }_{\hat{\theta} \hat{r}}.
(10.3) (a) Use u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1 to say
(c) A local observer measures time intervals d hat(t)=\mathrm{d} \hat{t}=(e_(t))^(t)dt=sqrt(-g_(tt))dt\left(\boldsymbol{e}_{t}\right)^{t} \mathrm{~d} t=\sqrt{-g_{t t}} \mathrm{~d} t and radial intervals dr=(e_(r))^( hat(r))dr\mathrm{d} r=\left(\boldsymbol{e}_{r}\right)^{\hat{r}} \mathrm{~d} r, so that we have
(10.5) Observers at both the emission and observation points are at rest in the rotating frame, so will have velocity with non-zero component u^(0)u^{0} given by
Since both emitter and observer are at the same value of rr, then they both find the same value of E=-p*uE=-\boldsymbol{p} \cdot \boldsymbol{u} for a photon of momentum p\boldsymbol{p}.
(11.2) (a)(i) R_(rrt)^(t)=(2M)/(r^(2)(2M-r))R_{r r t}^{t}=\frac{2 M}{r^{2}(2 M-r)}.
(ii) We have that R_(rtrt)=R_(trtr)=g_(tt)R^(t)_(rtr)R_{r t r t}=R_{t r t r}=g_{t t} R^{t}{ }_{r t r}, and so
R_(trt)^(r)=g^(rr)g_(tt)R_(rtr)^(t)=(2M(2M-r))/(r^(4))R_{t r t}^{r}=g^{r r} g_{t t} R_{r t r}^{t}=\frac{2 M(2 M-r)}{r^{4}}
(12.1) The transformation is carried out using the transformation components
{:[Lambda^(x)_(r)=sin theta cos phi",",Lambda^(x)_(theta)=r cos theta cos phi","],[Lambda^(x)=-r sin theta sin phi",",Lambda^(y)=sin theta sin phi","],[Lambda_(theta)^(y)=r cos theta sin phi",",Lambda_(phi)^(y)=r sin theta cos phi],[Lambda_(r)^(z)=cos theta",",Lambda^(z)_(theta)=-r sin theta],[Lambda^(z)_(phi)=0]:}\begin{array}{cc}
\Lambda^{x}{ }_{r}=\sin \theta \cos \phi, & \Lambda^{x}{ }_{\theta}=r \cos \theta \cos \phi, \\
\Lambda^{x}=-r \sin \theta \sin \phi, & \Lambda^{y}=\sin \theta \sin \phi, \\
\Lambda_{\theta}^{y}=r \cos \theta \sin \phi, & \Lambda_{\phi}^{y}=r \sin \theta \cos \phi \\
\Lambda_{r}^{z}=\cos \theta, & \Lambda^{z}{ }_{\theta}=-r \sin \theta \\
\Lambda^{z}{ }_{\phi}=0
\end{array}
Using these we find T_(tt)=rho,quadT_(rr)=p,quadT_(theta theta)=pr^(2),quadT_(phi phi)=pr^(2)sin^(2)thetaT_{t t}=\rho, \quad T_{r r}=p, \quad T_{\theta \theta}=p r^{2}, \quad T_{\phi \phi}=p r^{2} \sin ^{2} \theta.
(12.3) To apply the rule given in the question, write T^(mu nu)(x)=sum_(n)intdtau mz^(˙)_(n)^(mu)z^(˙)_(n)^(nu)delta^((3))[x-z_(n)(tau)]delta[x^(0)-z_(n)^(0)(tau)]T^{\mu \nu}(x)=\sum_{n} \int \mathrm{~d} \tau m \dot{z}_{n}^{\mu} \dot{z}_{n}^{\nu} \delta^{(3)}\left[x-z_{n}(\tau)\right] \delta\left[x^{0}-z_{n}^{0}(\tau)\right], (E.101) then take f(tau_(n))=x^(0)-z_(n)^(0)(tau_(n))f\left(\tau_{n}\right)=x^{0}-z_{n}^{0}\left(\tau_{n}\right), with zeros at tau_(n)\tau_{n} when x^(0)=z^(0)(tau_(n))x^{0}=z^{0}\left(\tau_{n}\right) and derivative |f^(')(tau_(n))|=|dz_(n)^(0)(tau_(n))//dtau_(n)|\left|f^{\prime}\left(\tau_{n}\right)\right|=\left|\mathrm{d} z_{n}^{0}\left(\tau_{n}\right) / \mathrm{d} \tau_{n}\right|. Doing the integral, we obtain an energy-momentum tensor T^(mu nu)(x)=sum_(n)m(z^(˙)_(n)^(mu)(tau_(n))z^(˙)_(n)^(nu)(tau_(n)))/(|(dz_(n)^(0))/((d)tau_(n))|)delta^((3))[x-z_(n)(tau_(n))]T^{\mu \nu}(x)=\sum_{n} m \frac{\dot{z}_{n}^{\mu}\left(\tau_{n}\right) \dot{z}_{n}^{\nu}\left(\tau_{n}\right)}{\left|\frac{\mathrm{d} z_{n}^{0}}{\mathrm{~d} \tau_{n}}\right|} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right]. (E.102)
(a) Set mu=alpha\mu=\alpha and nu=0\nu=0 and we have
where we've written p_(n)^(mu)(tau_(n))=mz^(˙)_(n)^(mu)(tau_(n))p_{n}^{\mu}\left(\tau_{n}\right)=m \dot{z}_{n}^{\mu}\left(\tau_{n}\right).
(b) Set mu=alpha\mu=\alpha and nu=i\nu=i and we have
and the answer follows from taking z^(0)(tau_(n))=tz^{0}\left(\tau_{n}\right)=t.
(c) Use the fact that mdz_(n)^(0)//dtau_(n)m \mathrm{~d} z_{n}^{0} / \mathrm{d} \tau_{n} is equal to the energy E_(n)(tau_(n))E_{n}\left(\tau_{n}\right) to write
Use the identity (1)/(sqrt(-g))(delsqrt(-g))/(delx^(mu))=Gamma^(alpha)_(mu alpha)\frac{1}{\sqrt{-g}} \frac{\partial \sqrt{-g}}{\partial x^{\mu}}=\Gamma^{\alpha}{ }_{\mu \alpha}, to find that the second term becomes -Gamma^(alpha)_(mu alpha)T^(mu nu)-\Gamma^{\alpha}{ }_{\mu \alpha} T^{\mu \nu} which, on relabelling indices, will cancel against the second term in eqn E.106. Since the delta function depends only on the separaSince the delta function depends only on the separa-
tion of xx and zz, we are allowed to replace del//delx^(mu)\partial / \partial x^{\mu} with tion of xx and zz, we are allowed to replace del//delx^(mu)\partial / \partial x^{\mu} with -del//delz^(mu)-\partial / \partial z^{\mu} in the first term. We also note that the chain -del//delz^(mu)-\partial / \partial z^{\mu} in the first term
rule allows us to replace
(E.112)
(b) In an orthonormal basis with v=e_( hat(0))v=\boldsymbol{e}_{\hat{0}}, we have T(,e_( hat(0)))=-rhoe_( hat(0))\boldsymbol{T}\left(, \boldsymbol{e}_{\hat{0}}\right)=-\rho \boldsymbol{e}_{\hat{0}} and T(,e_( hat(i)))=pe_( hat(i))\boldsymbol{T}\left(, \boldsymbol{e}_{\hat{i}}\right)=p \boldsymbol{e}_{\hat{i}}. As a consequence
(c) The right-hand side of the last equation is a number that can be rewritten as X^( hat(0))Y^( hat(0))(rho+p)+eta_( hat(mu) hat(nu))X^( hat(mu))Y^( hat(mu))pX^{\hat{0}} Y^{\hat{0}}(\rho+p)+\eta_{\hat{\mu} \hat{\nu}} X^{\hat{\mu}} Y^{\hat{\mu}} p, since Y_( hat(0))=-Y^(0)Y_{\hat{0}}=-Y^{0} in flat space. Hence we can extract a (0,2)(0,2) version of the tensor components. Using the fact that v=e_( hat(0))\boldsymbol{v}=\boldsymbol{e}_{\hat{0}}, we have that tilde(v)\tilde{\boldsymbol{v}} can be used to extract the timelike components of vectors. The corresponding flattimelike components of vectors. The corresponding flat-
space expression for T(Y,X)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X}) can then be reexpressed space expression for T(Y,X)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X}) can then be reexp
in vector notation using tilde(v)\tilde{\boldsymbol{v}} and the metric eta\boldsymbol{\eta} as T(Y,X)=(rho+p)[ tilde(v)(Y) tilde(v)(X)]+p eta(Y,X).(E.115)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X})=(\rho+p)[\tilde{\boldsymbol{v}}(\boldsymbol{Y}) \tilde{\boldsymbol{v}}(\boldsymbol{X})]+p \boldsymbol{\eta}(\boldsymbol{Y}, \boldsymbol{X}) .(\mathrm{E} .115)
In curved space, we swap eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}. We therefore suggest that the form of the (0,2)(0,2) tensor is
(15.1) Volume of a cylinder is pir^(2)h~~10^(11)\pi r^{2} h \approx 10^{11} in cubic parsecs, which is then the number of stars.
(15.2) (a) The invariant interval can be written in terms of the metric line element as
These can be shifted into the orthonormal frame using the vielbein given in Chapter 15.
(15.3) (a) The Einstein tensor for the state of affairs described is given by
{:(E.123)G_(mu nu)=R_(mu nu)-(1)/(2)Rg_(mu nu)=-(1)/(4)Rg_(mu nu):}\begin{equation*}
G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=-\frac{1}{4} R g_{\mu \nu} \tag{E.123}
\end{equation*}
(b) The components G_(mu nu)G_{\mu \nu} must be equal to the energymomentum contribution (=-Lambdag_(mu nu))\left(=-\Lambda g_{\mu \nu}\right), so that the Einstein equation reads
So fixing a Ricci scalar R=4LambdaR=4 \Lambda gives us the same flat, expanding Universe that we had before.
(15.5) (a) The energy density is of order 1GeV//(10^(-45)m^(3))~~1 \mathrm{GeV} /\left(10^{-45} \mathrm{~m}^{3}\right) \approx1.602 xx10^(35)Jm^(-3)1.602 \times 10^{35} \mathrm{Jm}^{-3}.
(b) In SI units, the Einstein equation is
where the factor (8pi G)/(c^(4))=2.07 xx10^(-43)m^(-1)kg^(-1)s^(2)\frac{8 \pi G}{c^{4}}=2.07 \times 10^{-43} \mathrm{~m}^{-1} \mathrm{~kg}^{-1} \mathrm{~s}^{2}. The right-hand-side is therefore ~~3xx10^(-8)m^(-2)\approx 3 \times 10^{-8} \mathrm{~m}^{-2}. Take the left-hand side to be equal to the curvature 1//a^(2)1 / a^{2}, where aa is the characteristic size of the Universe, we have a~~5kma \approx 5 \mathrm{~km}. We conclude that the vacuum energy of the strong nuclear force would wind the Universe into a ball with a radius of a few km . We could barely go for a walk before finding we are back where we started!
(15.6) (a) The equation of motion for tt is
The rr-dependence can be eliminated by substituting using the length parametrization condition L=1L=1.
(c) Substituting into the velocity identity g_(mu nu)u^(mu)u^(nu)=g_{\mu \nu} u^{\mu} u^{\nu}= -1 , we find
If we take g_(ii)(u^(i))^(2)=| vec(u)|^(2)g_{i i}\left(u^{i}\right)^{2}=|\vec{u}|^{2} then using the solution of the equation of motion to substitute for dt//dtau\mathrm{d} t / \mathrm{d} \tau yields the answer.
(15.7) (a) Take ds=0\mathrm{d} s=0 and write dt//a(t)=dr\mathrm{d} t / a(t)=\mathrm{d} r, or
{:(E.130)int_(0)^(ℓ)dr=int_(t_(s))^(t_(r))dte^(-Ht):}\begin{equation*}
\int_{0}^{\ell} \mathrm{d} r=\int_{t_{\mathrm{s}}}^{t_{\mathrm{r}}} \mathrm{~d} t \mathrm{e}^{-H t} \tag{E.130}
\end{equation*}
from which the answer follows.
(b) As ℓ\ell increases, t_(r)t_{\mathrm{r}} also increases. The distance ℓ^(')=e^(-Ht_(s))//H\ell^{\prime}=\mathrm{e}^{-H t_{\mathrm{s}}} / H corresponds to an infinite receipt time t_(r)rarr oot_{\mathrm{r}} \rightarrow \infty. Therefore, an observer beyond a radius ℓ^(')\ell^{\prime} will never receive the light signal sent from the origin, since the expansion of the Universe is fast enough to prevent this. Notice this scale is set by 1//H1 / H.
(15.8) From Exercise 15.2, we start from one of the equations of motion, in the form,
where we've used the compatibility condition for the derivatives of the metric components (i.e. g_(mu nu;sigma)=0g_{\mu \nu ; \sigma}=0 ). Next, contract indices by (i) multiplying by g^(mu alpha)g^{\mu \alpha} and summing, then (ii) multiplying by g^(nu beta)g^{\nu \beta} and summing. We find that C_(,sigma)=0C_{, \sigma}=0, so CC must be constant.
(16.6) We write
and so R=12 CR=12 C. This means that R_(nu beta)=(1)/(4)Rg_(nu beta)R_{\nu \beta}=\frac{1}{4} R g_{\nu \beta} and also that the components of the Einstein tensor are G_(nu beta)=-(1)/(4)Rg_(nu beta)G_{\nu \beta}=-\frac{1}{4} R g_{\nu \beta}. In general, we have for dd-dimensional spacetime that ^((d))R_(mu nu)=(d-1)Cg_(mu nu){ }^{(d)} R_{\mu \nu}=(d-1) C g_{\mu \nu} and ^((d))R={ }^{(d)} R=Cd(d-1)C d(d-1). This allows us to write ^((d))R_(mu nu)=(^((d))R)/(d)g_(mu nu){ }^{(d)} R_{\mu \nu}=\frac{{ }^{(d)} R}{d} g_{\mu \nu}. (16.7) We have
This can be used to write ^((3))R_(ij)=(1)/(3)^((3))Rgamma_(ij){ }^{(3)} R_{i j}=\frac{1}{3}{ }^{(3)} R \gamma_{i j}, in a similar fashion to the previous question.
(16.8) We have that ℓ_(ob)//ℓ_(em)=a(t_(ob))//a(t_(em))\ell_{\mathrm{ob}} / \ell_{\mathrm{em}}=a\left(t_{\mathrm{ob}}\right) / a\left(t_{\mathrm{em}}\right). Since a(t_(ob))//a(t_(em))=(1+z)a\left(t_{\mathrm{ob}}\right) / a\left(t_{\mathrm{em}}\right)=(1+z), we have
(17.2) (a) Using U=-del(ln Z)//del betaU=-\partial(\ln Z) / \partial \beta, with beta=1//k_(B)T\beta=1 / k_{\mathrm{B}} T, we find that the internal energy per particle is 3k_(B)T3 k_{\mathrm{B}} T. (b) The pressure is found from p=-(del F//del V)_(T)p=-(\partial F / \partial V)_{T}, where F=-(1)/(beta)ln ZF=-\frac{1}{\beta} \ln Z is the free energy. We obtain a pressure k_(B)T//Vk_{\mathrm{B}} T / V per particle. The result in (c) follows from taking the ratio of these two results.
(17.3) Setting k=0k=0 we find
and so ds^(2)=-dT^(2)+dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right),
, (E.149)
which is the line element for empty Minkowski space (Universe 0).
(18.5) We have dX=TdT//X\mathrm{d} X=T \mathrm{~d} T / X, so
Writing T=sinh psiT=\sinh \psi, which implies X=cosh psiX=\cosh \psi, we find ds^(2)=-dpsi^(2)\mathrm{d} s^{2}=-\mathrm{d} \psi^{2}.
Usually we would call psi\psi the proper time tau\tau. The embedding of a hyperbolic world line in Minkowski space therefore tells us that the intervals along the world line can be measured with the proper time, which is something we already knew.
(18.6) (a) Eliminate YY to find
which simplifies to the expression in the question.
(b) Set v=sinh chiv=\sinh \chi to find that the relationship is
{:[T=sinh chi cosh psi","],[X=sinh chi sinh psi","],[(E.153)Y=cosh chi]:}\begin{align*}
T & =\sinh \chi \cosh \psi, \\
X & =\sinh \chi \sinh \psi, \\
Y & =\cosh \chi \tag{E.153}
\end{align*}
(c) The line element ds^(2)=-dchi^(2)+cosh^(2)chidtheta^(2)\mathrm{d} s^{2}=-\mathrm{d} \chi^{2}+\cosh ^{2} \chi \mathrm{~d} \theta^{2} is based on coordinates
{:[T=sinh chi],[X=cosh chi cos theta],[(E.154)Y=cosh chi sin theta]:}\begin{align*}
T & =\sinh \chi \\
X & =\cosh \chi \cos \theta \\
Y & =\cosh \chi \sin \theta \tag{E.154}
\end{align*}
This uses a set of circles in the X-YX-Y plane as the basis for the coordinate system used. So if we fix the timelike variable chi\chi, the coordinates describe a circle in the X-YX-Y plane. In contrast, a fixed time chi\chi in the other form of the line element gives us coordinates describing the hyperbolic curve -T^(2)+X^(2)=1-T^{2}+X^{2}=1 in the X-TX-T plane. perbolic curve -T^(2)+X^(2)=1-T^{2}+X^{2}=1 in the X-TX-T plane.
(d) In the absence of matter, we have a^(˙)^(2)+k=Lambdaa^(2)//3\dot{a}^{2}+k=\Lambda a^{2} / 3. (d) In the absence of matter, we have a^(˙)^(2)+k=Lambdaa^(2)//3\dot{a}^{2}+k=\Lambda a^{2} / 3.
In units where Lambda//3=1\Lambda / 3=1, we have that a(t)=cosh ta(t)=\cosh t corresponds to a closed (k=1)(k=1) universe, and a(t)=sinh ta(t)=\sinh t corresponds to an open universe.
(18.7) (b) Adding, we find T+X=e^(t)T+X=\mathrm{e}^{t} or t=ln(T+X)t=\ln (T+X). So the coordinates only cover the region T+X > 0T+X>0. (c) Lines of constant tt obey the equation T=A-XT=A-X, where AA is a constant. They also obey T=B+CY^(2)T=B+C Y^{2}, with BB and CC constants.
(d) The transformation needed here is u=e^(-t)u=\mathrm{e}^{-t}.
(18.8) The embedding only works for x < 1x<1. We have T+X=T+X=(1-x)^((1)/(2))e^(t) > 0(1-x)^{\frac{1}{2}} \mathrm{e}^{t}>0 and -T+X=(1-x)^((1)/(2))e^(-t) > 0-T+X=(1-x)^{\frac{1}{2}} \mathrm{e}^{-t}>0, so only the quarter of the hyperboloid with X >= |T|X \geq|T| is covered by these coordinates.
(19.2) (b) For beta=1\beta=1 we have du//dv=0\mathrm{d} u / \mathrm{d} v=0, that we might call 'at rest' in terms of the coordinate velocity. For beta=0\beta=0, the coordinate velocity du//dv=1\mathrm{d} u / \mathrm{d} v=1 ('light coordinate speed') and for beta=-1\beta=-1 we have infinite coordinate speed. In short, we should beware of interpretations of coordinate velocities.
(19.3) (a) In component form, we have
Substituting the transformation we find the factors of f(x)f(x) cancel, so the angle is unchanged.
(b) A null curve with tangent aa has g(a,a)=0g(a, a)=0. In component form, we have
which will still vanish, so the curve remains null.
(a) The form of the solution will be the same in the two spacetimes. The time dependence in terms of tt is therefore written as
(b) The instantaneous frequency omega(t)\omega(t) can be extracted by taking its derivative of the phase of the wave with respect to time tt, which tells us that omega(t)a(t)=\omega(t) a(t)= const.
(19.6) Putting the steps together we find
Fig. E. 1 Penrose diagram for Exercise 19.7 showing (a) dotted, (b) solid and (c) dashed.
(19.8) (a) Use a^(˙)^(2)+k=C//a+Lambdaa^(3)//3\dot{a}^{2}+k=C / a+\Lambda a^{3} / 3.
(b) Recall that a light ray travels a constant comoving distance
and integrate.
(e) The equations for a(eta)a(\eta) and t(eta)t(\eta) describe a cycloid for a(t)a(t). Starting at eta=0\eta=0, the Universe reaches its maximum radius a=Ca=C when eta=pi\eta=\pi, and then reaches the Big Crunch at eta=2pi\eta=2 \pi. The variable eta\eta has a convenient interpretation in terms of the angle over which light traverses the unit circle, so at eta=pi\eta=\pi we have light travelling halfway around the circle from each direction. At the moment of the big crunch, when eta=2pi\eta=2 \pi, light can therefore, only just, traverse the whole of the circle.
(20.3) (a) We have potential energy
(E.162)
where dM(r)\mathrm{d} M(r) is an element of mass separated from the particle by a distance rr. For the ring we have all of the elements of mass at a distance r=(a^(2)+x^(2))^((1)/(2))r=\left(a^{2}+x^{2}\right)^{\frac{1}{2}} from mm. We can at this point simply write the answer, or, if the mass per unit length is lambda=M//2pi a\lambda=M / 2 \pi a we can integrate elements of length adthetaa \mathrm{~d} \theta to find
U=-(GMm)/(2pi ar)int_(0)^(2pi)adtheta=-(GMm)/(r)=-(GMm)/((a^(2)+x^((1)/(2))_((E.163)).)U=-\frac{G M m}{2 \pi a r} \int_{0}^{2 \pi} a \mathrm{~d} \theta=-\frac{G M m}{r}=-\frac{G M m}{\left(a^{2}+\underset{(\mathrm{E} .163)}{x^{\frac{1}{2}}} .\right.}
(b) By symmetry, the force can only act along the xx direction (meaning F_(y)=F_(z)=0F_{y}=F_{z}=0 ), so we differentiate with respect to xx to find
{:(E.164)F_(x)=-GMm(x)/((a^(2)+x^(2))^((3)/(2))).:}\begin{equation*}
F_{x}=-G M m \frac{x}{\left(a^{2}+x^{2}\right)^{\frac{3}{2}}} . \tag{E.164}
\end{equation*}
In the limit that x≫ax \gg a, this becomes F_(x)=-GMm//x^(2)F_{x}=-G M m / x^{2}, the attractive force between two point masses.
(c) The disc may be built from concentric annular rings of radius aa, width da\mathrm{d} a and area 2pi ada2 \pi a \mathrm{~d} a. If we take the total mass per unit area of the disc to be sigma=Omega//(piL^(2))\sigma=\Omega /\left(\pi L^{2}\right), then each annular element has mass dOmega=2pi ada sigma\mathrm{d} \Omega=2 \pi a \mathrm{~d} a \sigma, and the contribution to the potential energy from each element is
Integrating and substituting for sigma\sigma yields
{:(E.166)U=-(2Gm Omega)/(L^(2))[(L^(2)+x^(2))^((1)/(2))-x]:}\begin{equation*}
U=-\frac{2 G m \Omega}{L^{2}}\left[\left(L^{2}+x^{2}\right)^{\frac{1}{2}}-x\right] \tag{E.166}
\end{equation*}
(20.4) (a) Multiply through by a velocity component u^(alpha)u^{\alpha} and sum to find
which, noting that u_(alpha)u^(alpha)=-1u_{\alpha} u^{\alpha}=-1, gives u^(beta)-u^(beta)=0u^{\beta}-u^{\beta}=0.
(20.5) The equation of motion can be rewritten as
whose solution is p^(alpha)e^(Phi)=p^{\alpha} e^{\Phi}= (const) for all of the coordinates. As a result, the ratios of the components of the momentum remain constant along the world line. There can therefore be no deflection of the light rays by the field.
(21.1) We find
(21.5) (a) We find g=r^(2)sin thetag=r^{2} \sin \theta and so the area is 4pir^(2)4 \pi r^{2}. (b) The circumference is 2pi r sin theta2 \pi r \sin \theta.
(21.6) (a) Consider a spherical mass of fluid and a sphericalshell mass element of the fluid at a radius rr, with surface shell mass element of the fluid at a radius rr, with surface
area AA, mass dm\mathrm{d} m and width dr\mathrm{d} r. There is an outwardarea AA, mass dm\mathrm{d} m and width dr\mathrm{d} r. There is an outward-
directed force on the inner surface of the element of p(r)Ap(r) A, resulting from the fluid closer to the centre. There are inward forces from (i) fluid outside the element, pushing back on the outer surface with force p(r+dr)Ap(r+\mathrm{d} r) A; and (ii) the gravitational pull of the fluid closer to the centre of Gm(r)dm//r^(2)\operatorname{Gm}(r) \mathrm{d} m / r^{2} (i.e. the fluid at a smaller radius acts as if it is all concentrated at the
origin in Newtonian gravitation). Equating forces we have
{:(E.172)Ap(r)=Ap(r+dr)+(Gm(r)dm)/(r^(2)):}\begin{equation*}
A p(r)=A p(r+\mathrm{d} r)+\frac{G m(r) \mathrm{d} m}{r^{2}} \tag{E.172}
\end{equation*}
Rearranging gives
{:(E.173)Adp=-(Gm(r)dm)/(r^(2)):}\begin{equation*}
A \mathrm{~d} p=-\frac{G m(r) \mathrm{d} m}{r^{2}} \tag{E.173}
\end{equation*}
where dp=p(r+dr)-p(r)\mathrm{d} p=p(r+\mathrm{d} r)-p(r). Finally we note that A=4pir^(2)A=4 \pi r^{2} and dm=rho Adr\mathrm{d} m=\rho A \mathrm{~d} r, where rho\rho is the density and we write m(r)=Mm(r)=M.
(21.7) (b) Set G_( hat(r) hat(t))=0G_{\hat{r} \hat{t}}=0 to find Lambda_(,t)=0\Lambda_{, t}=0.
(c) Set G_( hat(t) hat(t))=0G_{\hat{t} \hat{t}}=0 to find
and choose C=-2MC=-2 M.
(d) Consider G_( hat(r) hat(r))-G_(i hat(t))=0G_{\hat{r} \hat{r}}-G_{i \hat{t}}=0.
(e) Rescale dt^(')=e^(f(t))dt\mathrm{d} t^{\prime}=\mathrm{e}^{f(t)} \mathrm{d} t and the Schwarzschild metric is recovered.
(21.8) (a) Compare eqns 21.12 and 21.17, writing rho=-p=zeta\rho=-p=\zeta. (b) Substitution reveals we have a solution if H^(2)=H^{2}=8pi zeta//38 \pi \zeta / 3, which is the result for the Hubble constant in the de Sitter Universe.
(22.1) (a) The velocity has components
where the connection is given by Gamma^(mu)_(00)=-(1)/(2)g_(00,sigma)g^(mu sigma)\Gamma^{\mu}{ }_{00}=-\frac{1}{2} g_{00, \sigma} g^{\mu \sigma}. We then have a^(0)=0a^{0}=0 and
(b) Plugging in the components of the Schwarzschild metric we obtain g_(00,1)=-2M//r^(2)g_{00,1}=-2 M / r^{2} and so a^(mu)=a^{\mu}=(0,M//r^(2),0,0)\left(0, M / r^{2}, 0,0\right). That is, despite not moving, the particle is accelerating outwards as it is not following a ticle is ac
geodesic.
geodesic.
(c) Recall from Chapter 2 that the square of the proper acceleration alpha^(2)\alpha^{2} is equal to a^(2)=(1-2M//r)^(-1)(M^(2)//r^(4))\boldsymbol{a}^{2}=(1-2 M / r)^{-1}\left(M^{2} / r^{4}\right), so alpha\alpha is given by
(b) Set theta=pi//2\theta=\pi / 2 and theta^(˙)=0\dot{\theta}=0. We obtain the following, in terms of the conserved quantities:
(i) For the tt coordinate
(c) Substituting r^(˙)\dot{r} from the effective-energy equation into the equation of motion for r^(¨)\ddot{r} gives a simplified equation of motion
{:(E.187)r^(¨)+(M)/(r^(2))+(3M tilde(L)^(2))/(r^(4))-( tilde(L)^(2))/(r^(3))=0:}\begin{equation*}
\ddot{r}+\frac{M}{r^{2}}+\frac{3 M \tilde{L}^{2}}{r^{4}}-\frac{\tilde{L}^{2}}{r^{3}}=0 \tag{E.187}
\end{equation*}
Differentiating the effective energy equation with respect to proper time gives an identical equation, so these are consistent.
(22.3) Using the conservation of p_(t)p_{t} along the geodesic, we can relate the values of p^(t)p^{t} at radius RR and at oo\infty by raising the index and writing g_(tt)g_{t t} on both sides
We are looking to write this in terms of measured values p^( hat(t))p^{\hat{t}} [where {:p^(t)=(e_( hat(t)))^(t)p^(t)]\left.p^{t}=\left(e_{\hat{t}}\right)^{t} p^{t}\right], so we use the vielbein component (e_( hat(t)))^(t)=(1-2M//r)^(-(1)/(2))\left(e_{\hat{t}}\right)^{t}=(1-2 M / r)^{-\frac{1}{2}} so we have
Setting p^( hat(t))(R)=ℏomega_(R)p^{\hat{t}}(R)=\hbar \omega_{R} and p^( hat(t))(oo)=ℏomega_(oo)p^{\hat{t}}(\infty)=\hbar \omega_{\infty}, the answer follows.
(22.5) (a) The escape velocity is defined such that the velocity u^(r)u^{r} at r=oor=\infty is zero. At r=oor=\infty we have tilde(E)=1\tilde{E}=1, which is then true along the geodesic. This E=1E=1, which is then true along the geodesic. This
gives u^(t)=(1-2M//r)^(-1)u^{t}=(1-2 M / r)^{-1}. The effective energy equation, or u^(mu)u_(mu)=-1u^{\mu} u_{\mu}=-1, then gives u^(r)=(2M//r)^((1)/(2))u^{r}=(2 M / r)^{\frac{1}{2}}. The coordinate velocity at r=Rr=R is then
This is the same as the non-relativistic result.
(b) The energy measured by an observer is E=-muE=-m \boldsymbol{u}u_("obs ")\boldsymbol{u}_{\text {obs }} or (more simply) p^( hat(t))=mu^( hat(t))=mu^(t)(e_(t))^( hat(t))p^{\hat{t}}=m u^{\hat{t}}=m u^{t}\left(\boldsymbol{e}_{t}\right)^{\hat{t}}. These give
(22.6) (a) Write the angular speed of the earth as omega=dphi//dt\omega=\mathrm{d} \phi / \mathrm{d} t and we obtain, restoring factors,
{:(E.193)dtau_(1)=(1-(2GM)/(c^(2)r)-(r^(2)omega^(2))/(c^(2)))^((1)/(2))dt:}\begin{equation*}
\mathrm{d} \tau_{1}=\left(1-\frac{2 G M}{c^{2} r}-\frac{r^{2} \omega^{2}}{c^{2}}\right)^{\frac{1}{2}} \mathrm{~d} t \tag{E.193}
\end{equation*}
The second and third terms are small and so the answer follows on expanding the bracket to first order.
(b) Make the substitution r rarr r+hr \rightarrow r+h and omega r rarr\omega r \rightarrowomega(r+h)+v\omega(r+h)+v and expand, noting that h≪rh \ll r.
(c) Restoring factors we have that
Take v=250ms^(-1),h=10^(4)mv=250 \mathrm{~ms}^{-1}, h=10^{4} \mathrm{~m} and omega=10^(-4)rads^(-1)\omega=10^{-4} \mathrm{rads}^{-1}, and note that GM//r^(2)~~10ms^(-2)G M / r^{2} \approx 10 \mathrm{~ms}^{-2}, to find Delta~~1xx10^(-12)\Delta \approx 1 \times 10^{-12} for eastward flight.
(d) For westward flight substitute v rarr-vv \rightarrow-v to find Delta~~-2xx10^(-12)\Delta \approx-2 \times 10^{-12}.
(22.7) The observer at rest has velocity u\boldsymbol{u} with components u^(mu)=(u^(t),0,0,0)u^{\mu}=\left(u^{t}, 0,0,0\right), from which we can use the constraint on velocity to evaluate
Since we are planning to take a dot product, we only need to evaluate the timelike component of the other observer's velocity v\boldsymbol{v}. This observer is free falling, and so we have that - tilde(E)=v_(t)(r_(2))=v_(t)(r_(1))-\tilde{E}=v_{t}\left(r_{2}\right)=v_{t}\left(r_{1}\right). Since the observer so
Since gamma(v_("rel "))=(1-v_("rel ")^(2))^(-(1)/(2))\gamma\left(v_{\text {rel }}\right)=\left(1-v_{\text {rel }}^{2}\right)^{-\frac{1}{2}}, the latter can be rearranged to access v_("rel ")v_{\text {rel }}.
(23.1) (a) We have dr=dtheta=0\mathrm{d} r=\mathrm{d} \theta=0, and so the increment of proper time taken along the orbit is
Substituting dphi^(2)=(v^(2))/(r^(2))dt^(2)\mathrm{d} \phi^{2}=\frac{v^{2}}{r^{2}} \mathrm{~d} t^{2} we find
dtau=+-[(1-(2M)/(r))-v^(2)]^((1)/(2))dt\mathrm{d} \tau= \pm\left[\left(1-\frac{2 M}{r}\right)-v^{2}\right]^{\frac{1}{2}} \mathrm{~d} t
Everything in the square bracket is constant, and the integral intdt\int \mathrm{d} t over a period TT yields intdt=2pi r//v\int \mathrm{d} t=2 \pi r / v, giving
(b) Using the definition of tilde(L)\tilde{L} and eqn 23.11 we find
(u^(t))^(2)= tilde(L)^(2)//Mr.\left(u^{t}\right)^{2}=\tilde{L}^{2} / M r .
(E.203)
Now use this with the definition of tilde(E)\tilde{E}, combined with eqn 23.5 to obtain
{:(E.204)dtau=(1-(3M)/(r))^((1)/(2))dt:}\begin{equation*}
\mathrm{d} \tau=\left(1-\frac{3 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} t \tag{E.204}
\end{equation*}
Kepler's law tells us Delta t\Delta t for an orbit is equal to 2pi(r^(3)//M)^((1)/(2))2 \pi\left(r^{3} / M\right)^{\frac{1}{2}}, from which we obtain the answer. Kepler's law also implies that v=(M//r)^((1)/(2))v=(M / r)^{\frac{1}{2}}, which shows the answers from (a) and (b) are compatible.
(23.2) The method for solving problems of this type is explained in the answer to Exercise 23.1. Computing the proper time from the line element, we find Delta tau=\Delta \tau=T(1-v^(2))^((1)/(2))T\left(1-v^{2}\right)^{\frac{1}{2}}, where TT is the period in coordinate time. We then integrate the equation for v=e^(t)Rdphi//dtv=\mathrm{e}^{t} R \mathrm{~d} \phi / \mathrm{d} t, to find
(23.3) A free particle follows a geodesic with constant tilde(E)=\tilde{E}= A free particle follows a geodesic (1-2M//r)u^(t)(1-2 M / r) u^{t}. We therefore have
We conclude that t=c tau+dt=c \tau+d, with cc and dd constant. Note that a similar argument applies to the phi\phi variable Note that a similar argument applie
for motion at a constant value of theta\theta.
(23.4) (a) We have u^(mu)=(C,0,0,omega)u^{\mu}=(C, 0,0, \omega) and, using u^(2)=-1\boldsymbol{u}^{2}=-1, we find
{:(E.207)C^(2)=((1+omega^(2)r^(2))/(1-2M//r_(0))):}\begin{equation*}
C^{2}=\left(\frac{1+\omega^{2} r^{2}}{1-2 M / r_{0}}\right) \tag{E.207}
\end{equation*}
(b) Taking a covariant derivative yields grad_(u)u=(Gamma^(mu)_(tt)C^(2)+Gamma^(mu)_(t phi)C omega+Gamma^(mu)_(phi phi)omega^(2))e_(mu)\boldsymbol{\nabla}_{u} \boldsymbol{u}=\left(\Gamma^{\mu}{ }_{t t} C^{2}+\Gamma^{\mu}{ }_{t \phi} C \omega+\Gamma^{\mu}{ }_{\phi \phi} \omega^{2}\right) \boldsymbol{e}_{\mu}. (E.208) The only connection coefficients we need are Gamma^(r)_(tt)=\Gamma^{r}{ }_{t t}=M(r_(0)-2M)//r_(0)^(3)M\left(r_{0}-2 M\right) / r_{0}^{3}, and Gamma_(phi phi)^(r)=-(r_(0)-2M)\Gamma_{\phi \phi}^{r}=-\left(r_{0}-2 M\right), giving
{:(E.209)a=((M)/(r_(0)^(2))+3Momega^(2)-r_(0)omega^(2))e_(r):}\begin{equation*}
\boldsymbol{a}=\left(\frac{M}{r_{0}^{2}}+3 M \omega^{2}-r_{0} \omega^{2}\right) \boldsymbol{e}_{r} \tag{E.209}
\end{equation*}
For a particle far from the source of the field we need M//r_(0)^(2)=r_(0)omega^(2)M / r_{0}^{2}=r_{0} \omega^{2}, i.e. the gravitational acceleration supplies the centripetal acceleration required to keep the particle in orbit. When this is the case the particle follows a geodesic and we have no acceleration.
(23.5) (b) Referring to the figure in the chapter, we cut out a wedge delta\delta such that
{:(E.211)(2pi-delta)R=2pi r:}\begin{equation*}
(2 \pi-\delta) R=2 \pi r \tag{E.211}
\end{equation*}
where r=R cos alphar=R \cos \alpha and tan alpha=dz//dr\tan \alpha=\mathrm{d} z / \mathrm{d} r. Treating alpha\alpha as a small angle, we have
where in the final term we have approximated r≫2Mr \gg 2 M. (d) One third of the effect is accounted from this contribution. As pointed out in Zee (2013), rubbersheet models showing a constant-time slice of the Schwarzschild geometry can be misleading in trying to understand the origin of gravitational effects in relativity.
(23.6) (a) We find
(c) In the limit of small u_(c)u_{c}, we have u_(c)~~M tilde(E)^(2)// tilde(L)^(2)u_{c} \approx M \tilde{E}^{2} / \tilde{L}^{2}. (d) Substituting, we obtain
{:(E.216)w^('')+(1-4Mu_(c))w~~0:}\begin{equation*}
w^{\prime \prime}+\left(1-4 M u_{\mathrm{c}}\right) w \approx 0 \tag{E.216}
\end{equation*}
This is a simple harmonic oscillator equation with characteristic frequency omega_(0)^(2)=1-4Mu_(c)\omega_{0}^{2}=1-4 M u_{\mathrm{c}}.
(e) Using the hint we have
However, we also have that alpha=4M//b\alpha=4 M / b is the deflection angle, where b=theta_(i)D_(ℓ)b=\theta_{\mathrm{i}} D_{\ell}, so that
{:(E.220)theta_(i)=theta_(s)+(4MD_(ℓs))/(theta_(i)D_(ℓ)D_(s)):}\begin{equation*}
\theta_{\mathrm{i}}=\theta_{\mathrm{s}}+\frac{4 M D_{\ell \mathrm{s}}}{\theta_{\mathrm{i}} D_{\ell} D_{\mathrm{s}}} \tag{E.220}
\end{equation*}
The answer then follows using the definition of theta_(E)\theta_{\mathrm{E}}.
(b) We have that D_(ℓs)D_{\ell \mathrm{s}} and D_(s)D_{\mathrm{s}} are both far larger than D_(ℓ)D_{\ell}, which gives theta_(E)^(2)~~4M//D_(ℓ)\theta_{\mathrm{E}}^{2} \approx 4 M / D_{\ell}. For the case where the source, lens, and observer are collinear, we have theta_(s)=0\theta_{\mathrm{s}}=0 and theta_(i)=theta_(E)\theta_{\mathrm{i}}=\theta_{\mathrm{E}}. The image is then a ring with this angular radius.
(24.4) (a) Write dchi^(2)=(dr^(2))/(1-kr^(2))\mathrm{d} \chi^{2}=\frac{\mathrm{d} r^{2}}{1-k r^{2}}. For a radial ray dphi=dtheta=0\mathrm{d} \phi=\mathrm{d} \theta=0 and the result follows.
(b) Metric coefficients are independent of chi\chi along the ray and so we have a Killing vector e_(chi)\boldsymbol{e}_{\chi} and so u_(chi)=Cu_{\chi}=C is conserved along the ray. Raising the index using g^(chi chi)g^{\chi \chi} gives
u^(chi)=C//a(t)^(2)u^{\chi}=C / a(t)^{2}
(E.221)
Since u^(chi)=(dchi)/(dlambda)u^{\chi}=\frac{\mathrm{d} \chi}{\mathrm{d} \mathrm{\lambda}}, where lambda\lambda is the affine parameter we have
{:(E.223)+2(1-v^(2))RdrdT+r^(2)dOmega^(2).:}\begin{equation*}
+2\left(1-v^{2}\right) R \mathrm{~d} r \mathrm{~d} T+r^{2} \mathrm{~d} \Omega^{2} . \tag{E.223}
\end{equation*}
(b) Setting the square bracket to equal unity yields R=v(1-v^(2))^(-1)R=v\left(1-v^{2}\right)^{-1}.
(d) A slice of space at a constant value of TT is flat (e) The velocity dr//dtau\mathrm{d} r / \mathrm{d} \tau is unchanged from the Schwarzschild value u^(r)=-(2M//r)^((1)/(2))u^{r}=-(2 M / r)^{\frac{1}{2}}. Making the substitution and rearranging we find that we also have
and so the speed appears subluminal.
(25.2) (a) We first want to evaluate the interval between two events at fixed rr, measured by falling clocks. The interval is measured by one clock that falls through rr and a second clock that falls through rr an interval dt\mathrm{d} t later. Each clock requires the same interval in tt to fall to rr, so the second clock must have been dropped dt\mathrm{d} t seconds after the first. The difference in dt^(˘)\mathrm{d} \breve{t} registered between the arrival of the clocks is simply the difference dt\mathrm{d} t between the clocks being dropped and so dt^(˘)=dt\mathrm{d} \breve{t}=\mathrm{d} t.
(b) We are in the same situation as examined in Chapter 22 for radially falling observers. We saw there that dr//dtau=-(2M//r)^((1)/(2))\mathrm{d} r / \mathrm{d} \tau=-(2 M / r)^{\frac{1}{2}} and
(c) Write dt_(d)=-dr//(dr//dt)\mathrm{d} t_{\mathrm{d}}=-\mathrm{d} r /(\mathrm{d} r / \mathrm{d} t) and substitute from the previous equation.
(d) Use eqn 22.31 to compute dtau\mathrm{d} \tau via the integral
(e) The total time delay is the sum of (i) the delay in drop time and (ii) the delay in proper time to reach their different radial positions at a time tt.
(f) We evaluate
{:(E.230)dt^(˘)=dt+((2M//r)^((1)/(2)))/(1-2M//r)dr:}\begin{equation*}
\mathrm{d} \breve{t}=\mathrm{d} t+\frac{(2 M / r)^{\frac{1}{2}}}{1-2 M / r} \mathrm{~d} r \tag{E.230}
\end{equation*}
Substituting for dt\mathrm{d} t in the Schwarzschild metric gives the desired answer.
(25.4) (a) The stationary observer has only one non-zero velocity component u^(t)u^{t}, determined by the equation g_(tt)(u^(t))^(2)=-1g_{t t}\left(u^{t}\right)^{2}=-1, giving u^(t)=(1-2M//r)^(-(1)/(2))u^{t}=(1-2 M / r)^{-\frac{1}{2}}. We can therefore write
{:(E.232)v^(t)=(1-v_(0)^(2))^(-(1)/(2))(1-2M//r)^(-(1)/(2)):}\begin{equation*}
v^{t}=\left(1-v_{0}^{2}\right)^{-\frac{1}{2}}(1-2 M / r)^{-\frac{1}{2}} \tag{E.232}
\end{equation*}
The only other non-zero component of v^(mu)v^{\mu} is v^(phi)v^{\phi} and so, using g(v,v)=-1\boldsymbol{g}(\boldsymbol{v}, \boldsymbol{v})=-1, we have
(c) Plugging in, we find that for r=4Mr=4 M we have tilde(E)=[2(1-v_(0)^(2))]^(-(1)/(2))\tilde{E}=\left[2\left(1-v_{0}^{2}\right)\right]^{-\frac{1}{2}} and therefore
This makes sense if the particle is deflected at rr, since we must have E=V_("eff ")(r)\mathcal{E}=V_{\text {eff }}(r) in order for the motion to be tangential at the deflection point. A value of v_(0)=1//sqrt2v_{0}=1 / \sqrt{2} gives E=V_("eff ")=0\mathcal{E}=V_{\text {eff }}=0 and also tilde(L)//M=4M\tilde{L} / M=4 M. From Fig. 23.1 we see that the maximum value of the tilde(L)//M=4M\tilde{L} / M=4 M curve is V_("eff ")=0V_{\text {eff }}=0, occurring exactly at r=4Mr=4 M. We conclude that the particle is, only just, deflected from this point. In fact, v_(0)=1//sqrt2v_{0}=1 / \sqrt{2} represents the minimum value of the relative velocity that allows the particle to escape the hole from this point.
(26.4) (a) Note that e_(t)=(del//del t)^(mu)\boldsymbol{e}_{t}=(\partial / \partial t)^{\mu} is a Killing vector, and so u_(t)u_{t} is a constant of the motion, just as we've had previously. The constant can be rewritten using the metric components as
which is also a constant of the motion.
(b) Substitute for xx and tt, set u=u= const. and integrate to get the expression for the affine parameter for outgoing geodesics.
(27.4) Along with the world-line component x^(1)(tau)=x_(0)x^{1}(\tau)=x_{0}, we use u^(2)=-1\boldsymbol{u}^{2}=-1 to compute g_(tt)(u^(t))^(2)=-1g_{t t}\left(u^{t}\right)^{2}=-1 and find that the velocity has components u^(mu)=(1//x_(0),0)u^{\mu}=\left(1 / x_{0}, 0\right). Taking a covariant derivative yields an acceleration with non-zero components
Using the connection coefficients from Exercise 9.4, we conclude a^(mu)=(0,1//x_(0))a^{\mu}=\left(0,1 / x_{0}\right). The proper acceleration alpha\alpha has the property alpha^(2)=a^(2)\alpha^{2}=\boldsymbol{a}^{2}, so we have alpha=1//x_(0)\alpha=1 / x_{0} too.
27.5) (a) Identify tau\tau and the observer's proper time and gamma=\gamma=(1-v^(2))^(-(1)/(2))\left(1-v^{2}\right)^{-\frac{1}{2}} and the result follows.
(b) Substituting into the exponential we obtain
It's worth noting that the instantaneous frequency can be extracted by operating on the phase with -idel//del tau-\mathrm{i} \partial / \partial \tau. (c) Repeating we obtain
Since r_(S)=2Mr_{\mathrm{S}}=2 M, we have A=16 piM^(2)A=16 \pi M^{2}.
(28.2) Use M=ℏ//(8pik_(B)T),A=16 piM^(2),S=Ak_(B)//(4ℏ)M=\hbar /\left(8 \pi k_{\mathrm{B}} T\right), A=16 \pi M^{2}, S=A k_{\mathrm{B}} /(4 \hbar), and the result follows. The negative sign results from the fact that when a Schwarzschild black hole accretes matter, its energy and mass increases, but its temperature decreases. Similarly, when the black hole evaporates, it radiates energy, its mass decreases, and its temperature increases. Hence the negative heat capacity.
(28.3) (a) The required metric components, in matrix form for the coordinates (V,r)(V, r), are
Acting on xi^(mu)\xi^{\mu} with this matrix gives xi_(nu)\xi_{\nu}.
(c) The motivation for the acrobatics in this problem is to use xi^(mu)=(1,0,0,0)\xi^{\mu}=(1,0,0,0) to simplify the expressions. Start by choosing sigma=V\sigma=V in -xi^(mu)xi_(mu)^(;sigma)=kappaxi^(sigma)-\xi^{\mu} \xi_{\mu}^{; \sigma}=\kappa \xi^{\sigma} and then note that only the mu=V\mu=V component survives in the contraction to obtain xi_(V)^(;V)=-kappa\xi_{V}{ }^{; V}=-\kappa. There is a rule for dealing more directly with this equation for the diverdealing more directly with this equation for fie divergence that
can write
Then using
(E.248)
we obtain the answer.
(28.4) (d) The angle psi\psi repeats after 2pi2 \pi radians so that psi=\psi=psi+2pi n\psi+2 \pi n, with nn an integer. Since psi=t_(E)//4M\psi=t_{\mathrm{E}} / 4 M, this gives the repeat period in imaginary time as Deltat_(E)=8pi M\Delta t_{\mathrm{E}}=8 \pi M, corresponding to a temperature
{:(E.249)k_(B)T=(ℏc^(3))/(8pi GM):}\begin{equation*}
k_{\mathrm{B}} T=\frac{\hbar c^{3}}{8 \pi G M} \tag{E.249}
\end{equation*}
(E.250)
where DD is the determinant of the matrix in the question. Some algebra reveals D=-Deltasin^(2)thetaD=-\Delta \sin ^{2} \theta, from which the desired equations follow straightforwardly.
(29.8) Using dtau^(2)=-ds^(2)\mathrm{d} \tau^{2}=-\mathrm{d} s^{2}, we find
{:[dtau^(2)=[(1-(2M)/(R))+(4Ma)/(R^(2))*v:}],[{:-(1+(a^(2))/(R^(2))+(2Ma^(2))/(R^(3)))v^(2)]dt^(2).]:}\begin{align*}
\mathrm{d} \tau^{2}= & {\left[\left(1-\frac{2 M}{R}\right)+\frac{4 M a}{R^{2}} \cdot v\right.} \\
& \left.-\left(1+\frac{a^{2}}{R^{2}}+\frac{2 M a^{2}}{R^{3}}\right) v^{2}\right] \mathrm{d} t^{2} .
\end{align*}
Since everything in the square braces is time independent, we have
\begin{align*}
\mathrm{d} \tau= & T
\end{aligned} \begin{aligned}
& \left(1-\frac{2 M}{R}\right)+\frac{4 M a}{R^{2}} \cdot v \\
& \left.-\left(1+\frac{a^{2}}{R^{2}}+\frac{2 M a^{2}}{R^{3}}\right) v^{2}\right]^{\frac{1}{2}} \tag{E.252}
\end{align*}
with T=2pi R//vT=2 \pi R / v.
(30.1) Use the equation dt//ds=kappa p\mathrm{d} \boldsymbol{t} / \mathrm{d} s=\kappa \boldsymbol{p} and, for a time tt, note that dt//ds=(dt//dt)(dt//ds)\mathrm{d} t / \mathrm{d} s=(\mathrm{d} t / \mathrm{d} t)(\mathrm{d} t / \mathrm{d} s), with ds//dt=1\mathrm{d} s / \mathrm{d} t=1 for motion at unit speed. We therefore have
{:(E.253)(dt)/((d)t)=kappa p:}\begin{equation*}
\frac{\mathrm{d} t}{\mathrm{~d} t}=\kappa p \tag{E.253}
\end{equation*}
which is Newton's second law for a unit mass with velocity t\boldsymbol{t}, subject to a force F=kappa p\boldsymbol{F}=\kappa \boldsymbol{p}, where p\boldsymbol{p} is a unit vector perpendicular to the velocity tt and hence also to the curve.
(30.2) Consider the path from pole to equator and back shown in Fig. 11.2, where the vector ends up 90^(@)90^{\circ} out from where it started. We have
(30.7) (a) A computation yields P^(mu)_(nu)P^{\mu}{ }_{\nu}.
(b) Compute bar(X)^(mu)n_(mu)\bar{X}^{\mu} n_{\mu}, to show it vanishes. This actually follows from P^(mu)n_(mu)=0P^{\mu} n_{\mu}=0.
(31.2) A suitable 1-form is tilde(W)()=f_(i)dx^(i)()\tilde{\boldsymbol{W}}()=f_{i} \boldsymbol{d} x^{i}(). Inserting vec(X)\vec{X} we find
For (a), this becomes (1)/(2)A^(mu nu)(T_(mu nu)+T_(nu mu))\frac{1}{2} A^{\mu \nu}\left(T_{\mu \nu}+T_{\nu \mu}\right) or A^(mu nu)T_((mu nu))A^{\mu \nu} T_{(\mu \nu)}. In the same way, for (b) we get A^(mu nu)T_([mu nu])A^{\mu \nu} T_{[\mu \nu]}.
(31.5) (b) For an antisymmetric tensor, the answer simplifies to
(31.6) Remove mention of the point xi^(mu)\xi^{\mu} on either side. Then interpret the combination d//dtau\mathrm{d} / \mathrm{d} \tau as a vector. This can be written as d//dtau=(dx^(beta)//dtau)(del//delx^(beta))\mathrm{d} / \mathrm{d} \tau=\left(\mathrm{d} x^{\beta} / \mathrm{d} \tau\right)\left(\partial / \partial x^{\beta}\right), interpreted as a set of components multiplied by a set of basis vectors, e_(beta)=del//delx^(beta):\boldsymbol{e}_{\beta}=\partial / \partial x^{\beta}:
(31.7) (b) Diagram (ii) represents A^(mu)_(alpha beta)B^(alpha)A^{\mu}{ }_{\alpha \beta} B^{\alpha}. Diagram (iii) represents A^(mu)_(alpha beta[gamma)C^(rho beta)_(sigma lambda])A^{\mu}{ }_{\alpha \beta[\gamma} C^{\rho \beta}{ }_{\sigma \lambda]}
(c) Briefly, (iv) represents grad_(mu)F^(mu nu)=J^(nu)\nabla_{\mu} F^{\mu \nu}=J^{\nu} and (v) is F_(mu nu;lambda)+F_(nu lambda;mu)+F_(lambda mu;nu)=0F_{\mu \nu ; \lambda}+F_{\nu \lambda ; \mu}+F_{\lambda \mu ; \nu}=0. See Chapter 42 for more detail.
(d) The diagram can be written as (grad_(alpha)grad_(beta)-grad_(beta)grad_(alpha))Z^(mu)=\left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{\nabla}_{\beta}-\boldsymbol{\nabla}_{\beta} \boldsymbol{\nabla}_{\alpha}\right) Z^{\mu}=R^(mu)_(nu alpha beta)Z^(nu)R^{\mu}{ }_{\nu \alpha \beta} Z^{\nu}. This expression is examined in Chapter 35.
(32.2) (a) The first equation can be written as
(b) The acceleration A=0\boldsymbol{A}=0 along a geodesic, so Fermi transport becomes equivalent to parallel transport.
(c) Consider the dot product a*b\boldsymbol{a} \cdot \boldsymbol{b},
This implies that all angular relationships between the vectors that are Fermi transported are maintained.
(d) Consider the dot product a*u\boldsymbol{a} \cdot \boldsymbol{u},
(E.270)
where we have used the facts that: A( bar(u))=0\boldsymbol{A}(\overline{\boldsymbol{u}})=0 and u(u)=-1\boldsymbol{u}(\boldsymbol{u})=-1.
(e) Ordinary parallel transport does not maintain the
angle between vectors, but Fermi transport does by virtue of the term -(A^^u)-(\boldsymbol{A} \wedge \boldsymbol{u}), which projects the vector after parallel transport so that it is perpendicular to tangent and relative angles between a,b\boldsymbol{a}, \boldsymbol{b} and c\boldsymbol{c} are maintained.
(33.1) Write
As a result the contraction (which is equivalent to the delta function) can go either side of the Lie derivative.
(33.4) If u\boldsymbol{u} and v\boldsymbol{v} are Killing fields vectors then
Since £_(u)£_(v)-£_(v)£_(u)=£_([u,v])£_{\boldsymbol{u}} £_{\boldsymbol{v}}-£_{\boldsymbol{v}} £_{\boldsymbol{u}}=£_{[\boldsymbol{u}, \boldsymbol{v}]}, then we must have
so the commutator is also a Killing vector.
(E.275)
(34.1) We won't assume the commutators vanish. We have that grad_(u+n)v-grad_(v)(u+n)=[u+n,v]=[u,v]+[n,v]\nabla_{u+n} v-\nabla_{v}(u+n)=[u+n, v]=[u, v]+[n, v]
=grad_(u)v-grad_(v)u+grad_(n)v-grad_(v)n=\nabla_{u} v-\nabla_{v} u+\nabla_{n} v-\nabla_{v} n
Expanding the second term on the left and rearranging, we have
grad_(u+n)v-grad_(v)u-grad_(v)n=grad_(u)v-grad_(v)u+grad_(n)v-grad_(v)n\nabla_{u+n} v-\nabla_{v} u-\nabla_{v} n=\nabla_{u} v-\nabla_{v} u+\nabla_{n} v-\nabla_{v} n
grad_(u+n)v=grad_(u)v+grad_(n)v\nabla_{u+n} v=\nabla_{u} v+\nabla_{n} v
This is a result that follows from the first rule for the covariant derivative given in the chapter.
(34.3) (a) We have
This quantity vanishes if u\boldsymbol{u} is the tangent field to a geodesic.
(b) We write u^(alpha)u_(sigma)Gamma^(sigma)_(alpha1)=u^(alpha)u^(sigma)Gamma_(sigma alpha1)u^{\alpha} u_{\sigma} \Gamma^{\sigma}{ }_{\alpha 1}=u^{\alpha} u^{\sigma} \Gamma_{\sigma \alpha 1}, then substitute for the connection coefficient in terms of the metric components and make use of the symmetry of the products. A slick method is to directly symmetrize to say
where, in going between lines, we've used g_(alpha beta,gamma)=g_{\alpha \beta, \gamma}=Gamma_(alpha beta gamma)+Gamma_(beta alpha gamma)\Gamma_{\alpha \beta \gamma}+\Gamma_{\beta \alpha \gamma} and the fact that gg is independent of x^(1)x^{1}. (34.6) (a) Combine the result from Example 34.8 with the general rule for forming inner products from eqn 32.21 .
(34.7) To relate the two parametrizations we write
We see that we recover the geodesic equation if the second term vanishes, which is the case if ss and lambda\lambda are linearly related.
(34.8) If the magnitude of u\boldsymbol{u} is constant, then g(u,u)\boldsymbol{g}(\boldsymbol{u}, \boldsymbol{u}) is constant along the curve, and we must have
where we have reindexed the second term in the sum and used g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0. This implies u*grad_(u)u=\boldsymbol{u} \cdot \nabla_{\boldsymbol{u}} \boldsymbol{u}=u^(alpha)u^(mu)_(;alpha)u_(mu)=0u^{\alpha} u^{\mu}{ }_{; \alpha} u_{\mu}=0, even though we don't have the geodesic condition u^(alpha)u^(mu)_(;alpha)=0u^{\alpha} u^{\mu}{ }_{; \alpha}=0.
(35.3) (a) In comma notation, we have
(b) Since we can write the second term del^(mu)del_(nu)A^(nu)\partial^{\mu} \partial_{\nu} A^{\nu} or del_(nu)del^(mu)A^(nu)\partial_{\nu} \partial^{\mu} A^{\nu} in flat space we can write
These will not yield the same answer in curved space, where a non-zero R\boldsymbol{R} tells us that the covariant derivatives don't commute.
(c) The second equation becomes
Inserting zeta^(mu)=lambdau^(mu)\zeta^{\mu}=\lambda u^{\mu} gives the quoted equation. Since Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta} at the origin can't be a function of the arbitrary directions u^(mu)u^{\mu}, the connection coefficients must vanish at this point.
(c) Reexpress the covariant derivative as
and then substitute n_(nu)^(mu)=lambdadelta_(nu)^(mu)n_{\nu}^{\mu}=\lambda \delta_{\nu}^{\mu} and the expression for the derivative of the connection coefficients.
(g) Permuting indices gives three expressions. Take one, subtract the second and add the third. Then use the symmetries of the Riemann tensor.
(35.6) (a) We can read off, using grad_(nu)e_(mu)=Gamma^(alpha)_(nu mu)e_(alpha)\nabla_{\nu} \boldsymbol{e}_{\mu}=\Gamma^{\alpha}{ }_{\nu \mu} \boldsymbol{e}_{\alpha}, that
where the | hat(theta) hat(r)||\hat{\theta} \hat{r}| notation fixes the order of these two variables based on the order in the wedge product in eqn E.304. Note also that since omega_( hat(theta))^( hat(r))=-omega_( hat(hat(r)))\boldsymbol{\omega}_{\hat{\theta}}^{\hat{r}}=-\boldsymbol{\omega}_{\hat{\hat{r}}}, so we also have
so we guess that omega_( hat(hat(x)))^( hat(hat(x)))propomega^( hat(x))\boldsymbol{\omega}_{\hat{\hat{x}}}^{\hat{\hat{x}}} \propto \boldsymbol{\omega}^{\hat{x}}. The exterior derivative of the first basis 1 -form is
and we identify omega^( hat(x))_( hat(r))=-omega^( hat(x))\boldsymbol{\omega}^{\hat{x}}{ }_{\hat{r}}=-\boldsymbol{\omega}^{\hat{x}} and omega^( hat(hat(r)))_( hat(x))=omega^( hat(x))\boldsymbol{\omega}^{\hat{\hat{r}}}{ }_{\hat{x}}=\boldsymbol{\omega}^{\hat{x}}. Taking another exterior derivative we have
We find that R^( hat(hat(r)))_( hat(x) hat(r) hat(x))=-1R^{\hat{\hat{r}}}{ }_{\hat{x} \hat{r} \hat{x}}=-1 and then that R^( hat(hat(r)))_( hat(r) hat(x) hat(r))=-1R^{\hat{\hat{r}}}{ }_{\hat{r} \hat{x} \hat{r}}=-1. We conclude that the components of the Ricci tensor are
{:(E.324){:[omega^( hat(t))=dt",",omega^( hat(chi))=ad chi],[omega^( hat(theta))=a sin chi d theta",",omega^( hat(phi))=a sin chi sin theta d phi]:}:}\begin{array}{ll}
\boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} t, & \boldsymbol{\omega}^{\hat{\chi}}=a \boldsymbol{d} \chi \\
\boldsymbol{\omega}^{\hat{\theta}}=a \sin \chi \boldsymbol{d} \theta, & \boldsymbol{\omega}^{\hat{\phi}}=a \sin \chi \sin \theta \boldsymbol{d} \phi \tag{E.324}
\end{array}
Since omega^( hat(t))=dt\boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} t we find domega^( hat(t))=0\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=0. We guess that omega^( hat(t))_( hat(k))prop\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}} \proptoomega^( hat(k))\boldsymbol{\omega}^{\hat{k}}, where k=theta,phik=\theta, \phi or chi\chi. (We also have omega_( hat(k))^( hat(t)_( hat(k)))=omega^( hat(k)). hat(t)\boldsymbol{\omega}_{\hat{k}}^{\hat{t}_{\hat{k}}}=\boldsymbol{\omega}^{\hat{k}} . \hat{t}.) Another exterior derivative
This fixes the constant of proportionality, so that we assume omega^( hat(t))_( hat(k))=(a^(˙)//a)omega^( hat(k))\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}}=(\dot{a} / a) \boldsymbol{\omega}^{\hat{k}} The next exterior derivative
where we've suspended the summation convention. Now we work out the Ricci tensor. Using R^( hat(i) hat(t) hat(i))i=-R_( hat(i) hat(i))^(i)R^{\hat{i} \hat{t} \hat{i}} \boldsymbol{i}=-R_{\hat{i} \hat{i}}^{i} (no summation implied), we find
We then have non-zero contributions to R_( hat(nu))^( hat(mu))\mathcal{R}_{\hat{\nu}}^{\hat{\mu}} of
and substitution of these gives the required results.
(37.1) (a) Insert the velocity and we have tilde(sigma)(u)=-Vu^(alpha)u_(alpha)=\tilde{\boldsymbol{\sigma}}(\boldsymbol{u})=-V u^{\alpha} u_{\alpha}= VV, as required for tilde(sigma)\tilde{\boldsymbol{\sigma}}.
(b) The total momentum in the box is
(37.2) (a) Since the box lives in the observer's three space, the tetrad ( u,A,B,C)\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}) spans spacetime.
(b) Inserting the velocity 1 -form into the current gives n=-J( tilde(u))n=-\boldsymbol{J}(\tilde{\boldsymbol{u}}). We can then write the current vector as
where alpha,beta\alpha, \beta and gamma\gamma are constants, then each box captures the same number of particles.
(37.5) (a) We have, by definition of the covariant derivative,
(37.6) (a) Trivially, £_(J)J=0£_{J} J=0. The other equalities follow from the fact that the box is rigid, so its corners must continue to be connected after they are transported along their geodesics.
(b) This follows from the result of the previous problem. (c) The quantity chi\chi is simply the number of particles in the box. Since this is a constant, by construction, we have £_(J)chi=0£_{J} \chi=0.
(d) Since theta_(J)=0\theta_{\boldsymbol{J}}=0, we also have from the previous problem that grad*J=0\boldsymbol{\nabla} \cdot \boldsymbol{J}=0.
(38.2) (a) If tilde(omega)\tilde{\boldsymbol{\omega}} is exact then tilde(omega)=d tilde(A)\tilde{\boldsymbol{\omega}}=\boldsymbol{d} \tilde{\boldsymbol{A}}, where tilde(A)\tilde{\boldsymbol{A}} is a 1-form. By Stokes theorem
since delS^(2)=0\partial S^{2}=0.
(b) We have d tilde(G)=dx^(1)^^dx^(2)^^dx^(3)\boldsymbol{d} \tilde{\boldsymbol{G}}=\boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}, which is the volume 3 -form. Taken over the volume VV of the ball from the last question, the integral
{:(E.361)int_(V)d tilde(G)!=0:}\begin{equation*}
\int_{V} d \tilde{G} \neq 0 \tag{E.361}
\end{equation*}
since it must give the volume inside the ball. By Stokes' theorem this tell us that
Given the result from (a), this implies that GG is not exact on S^(2)S^{2}. It is, however, closed on S^(2)S^{2}, as is the case for all 2 -forms, since all 3 -forms vanish in a two-dimensional space.
(38.4) (a) Using the results in Chapter 36 we find Gamma^( hat(phi))_( hat(phi) hat(theta))=cot theta\Gamma^{\hat{\phi}}{ }_{\hat{\phi} \hat{\theta}}=\cot \theta and Gamma_(phi^(˙) hat(phi))^( hat(phi))=-cot theta\Gamma_{\dot{\phi} \hat{\phi}}^{\hat{\phi}}=-\cot \theta. We then use the geodesic equation, noting that u^( hat(phi))=(e_(phi))^( hat(phi))u^(phi)=sin thetaphi^(˙)u^{\hat{\phi}}=\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} u^{\phi}=\sin \theta \dot{\phi}.
(c) A trick is needed here: we write the integral I\mathcal{I} as
which we interpret as taken around a path, parametrized by tau\tau, that forms a closed loop that bounds a surface SS. This allows us to use Green's theorem in the plane to write
Expanding the exponential exp M_=1+M_+M_^(2)//2!+\exp \underline{\boldsymbol{M}}=1+\underline{\boldsymbol{M}}+\underline{\boldsymbol{M}}^{2} / 2!+ ..., we find the rotation matrix
(39.1) Since rho gy+(1)/(2)rhov^(2)+p\rho g y+\frac{1}{2} \rho v^{2}+p (where yy is the height) is a constant, and the pressure is the same at the top of the reservoir as at the outlet. We have
{:(E.367)rho gy_(2)=rho gy_(1)+(1)/(2)*rhov^(2):}\begin{equation*}
\rho g y_{2}=\rho g y_{1}+\frac{1}{2} \cdot \rho v^{2} \tag{E.367}
\end{equation*}
since at the top of the reservoir (at height y_(2)y_{2} ) we assume the fluid has no velocity. We obtain v=[2g(y_(2)-y_(1))]^((1)/(2))v=\left[2 g\left(y_{2}-y_{1}\right)\right]^{\frac{1}{2}}, which is the same result for a particle having fallen through a distance y_(2)-y_(1)y_{2}-y_{1} in a constant gravitational field gg.
(39.2) (a) We find
where the final equality follows for the steady-flow condition (d)/(d tau)=u*grad=u^(0)(del)/(del t)+ vec(u)* vec(grad)=gamma vec(v)* vec(grad)\frac{d}{d \tau}=\boldsymbol{u} \cdot \nabla=u^{0} \frac{\partial}{\partial t}+\vec{u} \cdot \vec{\nabla}=\gamma \vec{v} \cdot \vec{\nabla}.
(39.3) (a) Start with the conservation equation (nu^(alpha)),alpha=0\left(n u^{\alpha}\right), \alpha=0, where we use commas rather than semicolons since we're working in flat spacetime. We obtain
Plugging in components yields the desired expression. (b) Start with the Euler equation in the form ( rho+\rho+ p) u_(alpha,nu)u^(nu)+p_(,alpha)+p_(,nu)u^(nu)u_(alpha)=0u_{\alpha, \nu} u^{\nu}+p_{, \alpha}+p_{, \nu} u^{\nu} u_{\alpha}=0, and repeat the same procedure as in part (a) to obtain the equation.
(c) If we have for a perfect fluid that
then we have an unchanging entropy to first order. Plugging into the first law of thermodynamics then yields the stated equation.
(d) Using the definition, we can write delta p=c_(s)^(2)delta rho\delta p=c_{\mathrm{s}}^{2} \delta \rho, so that the first law gives c_(s)^(2)(p_(**)+rho_(**))delta n//n_(**)=delta pc_{s}^{2}\left(p_{*}+\rho_{*}\right) \delta n / n_{*}=\delta p. Substituting into the result from (b) we find
Take the gradient of this latter expression and substitute into the result from (a) to obtain the wave equation.
(39.4) Using the identities given in the question we have grad*J_(m)=(m)/(sqrt(-g))intdtau((d)z^(alpha))/(dtau)(del)/(delx^(alpha))delta^((4))[x-z(tau)]\boldsymbol{\nabla} \cdot \boldsymbol{J}_{\mathrm{m}}=\frac{m}{\sqrt{-g}} \int \mathrm{~d} \tau \frac{\mathrm{~d} z^{\alpha}}{\mathrm{d} \tau} \frac{\partial}{\partial x^{\alpha}} \delta^{(4)}[x-z(\tau)]
(39.5) The only components of T\boldsymbol{T} we need are T^(tt)=rhoe^(-2Phi)T^{t t}=\rho \mathrm{e}^{-2 \Phi} and T^(rr)=pe^(-2Lambda)T^{r r}=p \mathrm{e}^{-2 \Lambda}. We consider the component equation T^(mu nu)_(;mu)=T^(mu nu)_(,mu)+Gamma^(mu)_(mu sigma)T^(sigma nu)+Gamma^(nu)_(mu sigma)T^(mu sigma).quadT^{\mu \nu}{ }_{; \mu}=T^{\mu \nu}{ }_{, \mu}+\Gamma^{\mu}{ }_{\mu \sigma} T^{\sigma \nu}+\Gamma^{\nu}{ }_{\mu \sigma} T^{\mu \sigma} . \quad (E.375)
First set mu=t\mu=t and nu=r\nu=r to find
Adding the two contributions and equating them to zero then gives the answer.
(39.6) (b) We first note that (:d tilde(omega), vec(v):)=0\langle\boldsymbol{d} \tilde{\omega}, \vec{v}\rangle=0. Then, plugging in, (40.4) Computing we find
we have
{:[" we have "],[£_( vec(v))(rho_(0)( tilde(omega)))],[=d(:( tilde(omega)),rho_(0)( vec(v)):)],[=d(rho_(0)v^(x)dy^^dz-rho_(0)v^(y)dx^^dz+rho_(0)v^(z)dx^^dy)],[=((del(rho_(0)v^(x)))/(del x)+(del(rho_(0)v^(y)))/(del y)+(del(rho_(0)v^(z)))/(del z))dx^^dy^^dz],[(E.378)=[( vec(grad))*(rho_(0)( vec(v)))] tilde(omega).]:}\begin{align*}
& \text { we have } \\
& £_{\vec{v}}\left(\rho_{0} \tilde{\boldsymbol{\omega}}\right) \\
&=\boldsymbol{d}\left\langle\tilde{\boldsymbol{\omega}}, \rho_{0} \vec{v}\right\rangle \\
&= \boldsymbol{d}\left(\rho_{0} v^{x} \boldsymbol{d} y \wedge \boldsymbol{d} z-\rho_{0} v^{y} \boldsymbol{d} x \wedge \boldsymbol{d} z+\rho_{0} v^{z} \boldsymbol{d} x \wedge \boldsymbol{d} y\right) \\
&=\left(\frac{\partial\left(\rho_{0} v^{x}\right)}{\partial x}+\frac{\partial\left(\rho_{0} v^{y}\right)}{\partial y}+\frac{\partial\left(\rho_{0} v^{z}\right)}{\partial z}\right) \boldsymbol{d} x \wedge \boldsymbol{d} y \wedge \boldsymbol{d} z \\
&= {\left[\vec{\nabla} \cdot\left(\rho_{0} \vec{v}\right)\right] \tilde{\boldsymbol{\omega}} . } \tag{E.378}
\end{align*}
(39.7) Since v_(j)=v^(j)v_{j}=v^{j} in flat space, we can write a 1 -form version of the given equation as
Substituting the other derivatives for (0,1)(0,1) exterior derivatives for consistency, we get the answer.
(39.8) A suitable 2-form is tilde(f)=epsi_(ijk)v^(i)[dx^(j)()^^dx^(k)()]\tilde{\boldsymbol{f}}=\varepsilon_{i j k} v^{i}\left[\boldsymbol{d} x^{j}() \wedge \boldsymbol{d} x^{k}()\right], where epsi_(ijk)\varepsilon_{i j k} is the three-dimensional antisymmetric symbol. When the two displacement vectors are inserted into the slots, we obtain the triple scalar product between vec(v), vec(a)\vec{v}, \vec{a} and vec(b)\vec{b}.
(40.1) (a) Using the Euler-Lagrange equation, the equation of motion is found to be
These can be used to derive the geodesic equation in the form given in (a), noting the minus sign going between deltag^(mu nu)\delta g^{\mu \nu} and deltag_(mu nu)\delta g_{\mu \nu}.
(d) A metric component with no dependence on x^(sigma)x^{\sigma} causes the right-hand side of the equation of motion to vanish, showing that momentum component p_(sigma)p_{\sigma} is conserved.
(40.2) (a) We find
(b) We can compare the previous expression against the rule for computing vec(grad)^(2)\vec{\nabla}^{2} in spherical polars, which is (del^(2))/(delr^(2))+(1)/(r^(2))*(del^(2))/(deltheta^(2))+(1)/(r^(2)sin^(2)theta)*(del^(2))/(delphi^(2))+(2)/(r)(del)/(del r)+(cos theta)/(r^(2)sin theta)(del)/((E.390))\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r^{2}} \cdot \frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2} \sin ^{2} \theta} \cdot \frac{\partial^{2}}{\partial \phi^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{\cos \theta}{r^{2} \sin \theta} \frac{\partial}{(\mathrm{E} .390)}.
This allows us to check that the spatial parts behave as expected when k=a^(˙)=0k=\dot{a}=0. If we are interested in slowly varying solutions in space, then the value of kk does not varying solut
(42.2) For the usual passive transformation using eqn 2.10 the components are
For an active transformation, reverse the sign of beta\beta.
(42.4) For both (a) and (b) we have vec(B)=C vec(e)_(z)\vec{B}=C \vec{e}_{z}.
(42.13) (a) We find
{:(E.392)L~~-mc^(2)+(1)/(2)mv^(2)-lambda phi+dots:}\begin{equation*}
L \approx-m c^{2}+\frac{1}{2} m v^{2}-\lambda \phi+\ldots \tag{E.392}
\end{equation*}
(c) Invoking the equivalence principle for the equation of motion must remove the dependence on mm. Scaling lambda\lambda and rescaling the field, we find that the equation of motion must take the form
The action then takes the form suggested in the question for consistency.
(42.14) (b) A scalar that describes both matter and energy is the trace TT of the energy-momentum tensor T\boldsymbol{T}. Using this we can upgrade the Lagrangian to read
This can be recreated from the contraction of the Bianchi identity R_(alpha[beta nu;mu])^(beta)=0R_{\alpha[\beta \nu ; \mu]}^{\beta}=0.
(43.3) Taking the dual with respect to the vector part of the (1,1)(1,1) tensor (and raising an index) will result in a (1,3)(1,3) tensor ***T=(***T)^(nu)_(alpha beta gamma)(e_(nu)oxomega^(alpha)^^omega^(beta)^^omega^(gamma))\star \boldsymbol{T}=(\star T)^{\nu}{ }_{\alpha \beta \gamma}\left(\boldsymbol{e}_{\nu} \otimes \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} \wedge \boldsymbol{\omega}^{\gamma}\right) with components
The expressions for the components of ***G\star \boldsymbol{G} are similar.
(43.4) Apply the operator d\boldsymbol{d} to T^(mu nu)(e_(mu)oxdSigma_(nu))T^{\mu \nu}\left(\boldsymbol{e}_{\mu} \otimes \mathrm{d} \boldsymbol{\Sigma}_{\nu}\right). We obtain
the second is sqrt(-g)T^(mu nu)de_(mu)=sqrt(-g)Gamma_(lambda sigma)^(mu)T^(sigma lambda)e_(mu)ox dx^(nu),quad\sqrt{-g} T^{\mu \nu} \boldsymbol{d} e_{\mu}=\sqrt{-g} \Gamma_{\lambda \sigma}^{\mu} T^{\sigma \lambda} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}, \quad (E.399) and the final one is
So we obtain T^(mu lambda)_(;lambda)e_(mu)ox dx^(nu)T^{\mu \lambda}{ }_{; \lambda} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}. Consider the wedge product: the only parts that survive will have nu\nu distinct from alpha,beta\alpha, \beta and gamma\gamma. We therefore obtain
The components are all zero by the usual conservation law T^(mu nu)_(;nu)=0T^{\mu \nu}{ }_{; \nu}=0.
(43.5) We run the argument from the chapter. The conservation law, written as a
{:(E.402)int_(V)d***T=0:}\begin{equation*}
\int_{\mathcal{V}} d \star T=0 \tag{E.402}
\end{equation*}
where DeltaS^(mu nu)=deltax^(mu)Deltax^(nu)\Delta S^{\mu \nu}=\delta x^{\mu} \Delta x^{\nu} is the area ABCD\mathcal{A B C D}. (Here the subscripts give the point where the field is evaluated along with a number denoting which step we're doing in the computation.)
(b) The result here is simply
where V^(rho mu nu)=dx^(rho)deltax^(mu)Deltax^(nu)V^{\rho \mu \nu}=\mathrm{d} x^{\rho} \delta x^{\mu} \Delta x^{\nu} is the volume of the cube. Since the differential operator acts on G_(mu nu)G_{\mu \nu} and also psi_(A0)\psi_{\mathcal{A} 0}, we may write, approximately, that
{:(E.410)psi_(A4)=(1-ig DeltaV^(rho mu nu)D_(rho)G_(mu nu))psi_(A0):}\begin{equation*}
\psi_{\mathcal{A} 4}=\left(1-\mathrm{i} g \Delta V^{\rho \mu \nu} D_{\rho} G_{\mu \nu}\right) \psi_{\mathcal{A} 0} \tag{E.410}
\end{equation*}
(g) The argument in Chapter 43 tells us that traversing the path makes zero contribution to the field, so the bracket in eqn 44.26 can be set to zero. Substitute for G_(mu nu)G_{\mu \nu} and confirm that the expressions are identical.
(44.2) The symmetrized energy-momentum tensor has components
(45.3) Use the fact that T_(mu nu)=(rho+p)u_(mu)u_(nu)+g_(mu nu)pT_{\mu \nu}=(\rho+p) u_{\mu} u_{\nu}+g_{\mu \nu} p to find the following:
(a) rho_(g)=2rho+(-rho+3p)\rho_{\mathrm{g}}=2 \rho+(-\rho+3 p).
(b) Pi_(i)=-(rho+p)u_(i)u_(0)\Pi_{i}=-(\rho+p) u_{i} u_{0} where, at low velocity, we have u^(mu)=(1, vec(u))u^{\mu}=(1, \vec{u}) and u_(mu)=(-1, vec(u))u_{\mu}=(-1, \vec{u}).
(c) rho_(c)=2p-(-rho+3p)\rho_{\mathrm{c}}=2 p-(-\rho+3 p).
(45.4) The key here is that we only retain terms up to order vec(u)^(i)\vec{u}^{i}, ignoring those at order u^(2)u^{2}. In the weak-field, lowvelocity limit we have a geodesic equation
If, in the absence of matter, we were to take the effective energy density to be T_(mu nu)^(')=-(Lambda)/(8pi)eta_(mu nu)T_{\mu \nu}^{\prime}=-\frac{\Lambda}{8 \pi} \eta_{\mu \nu}, then we have a trace T^(')=-Lambda//2piT^{\prime}=-\Lambda / 2 \pi and find an energy density rho_(g)=2T_(tt)^(')-eta_(tt)T^(')=-Lambda//4pi\rho_{\mathrm{g}}=2 T_{t t}^{\prime}-\eta_{t t} T^{\prime}=-\Lambda / 4 \pi. This is negative and so the force is repulsive.
(b) We find a field equation vec(grad)^(2)h_(tt)=2Lambda\vec{\nabla}^{2} h_{t t}=2 \Lambda and h_(tt)=-2Phih_{t t}=-2 \Phi, giving vec(grad)^(2)Phi=-Lambda\vec{\nabla}^{2} \Phi=-\Lambda. The function Phi=alphar^(2)\Phi=\alpha r^{2} solves the resulting equation
for alpha=-Lambda//6\alpha=-\Lambda / 6.
(45.6) (b) Using the usual formula for Gamma_(mu nu sigma)\Gamma_{\mu \nu \sigma} in terms of the components g_(mu nu)g_{\mu \nu}, we find
To the order at which we're working, raising the index can be done with the Minkowski tensor. (Alternatively, use g^(mu nu)g^{\mu \nu} and drop higher order terms.)
(45.7) (a) Use S_(mu)u^(mu)=0S_{\mu} u^{\mu}=0 to say
(45.8) Parallel transport preserves the magnitude of a vector, so we can say that g^(mu nu)S_(mu)S_(nu)g^{\mu \nu} S_{\mu} S_{\nu} is constant, which means
Then use S_(0)=-(dx^(i)//dt)S_(i)S_{0}=-\left(\mathrm{d} x^{i} / \mathrm{d} t\right) S_{i} and v^(i)=dx^(i)//dtv^{i}=\mathrm{d} x^{i} / \mathrm{d} t and ignore the non-diagonal term with g^(ij)g^{i j}, since this is higher order. Note also that in writing the 3 -vector, indices can der. Note also that in writing the 3 -vector, indices can
be raised and lowered with the Minkowski metric tenbe raised and lo
sor, so v_(i)=v^(i)v_{i}=v^{i}.
(45.9) (a) Dot the equation for vec(S)\vec{S} with itself.
(b) Compute vec(v)* vec(S)\vec{v} \cdot \vec{S}.
(c) Combine (a) and (b) with eqn 45.60
(e) Taking a derivative, we have
Using the rule for a double vector product, we can spot that the final two terms are equal to (3)/(2)[ vec(S)xx( vec(v)xx vec(grad)phi]\frac{3}{2}[\vec{S} \times(\vec{v} \times \vec{\nabla} \phi] and the answer follows.
(f) First, vec(grad)phi=M vec(r)//r^(3)\vec{\nabla} \phi=M \vec{r} / r^{3}. Then use components to say
{:[( vec(grad)xx zeta)_(a)=sum_(bc)epsi_(abc)grad_(b)zeta_(c)],[=sum_(bckm)epsi_(abc)(del)/(delx^(b))(1)/(r^(3))epsi_(ckm)x^(k)J^(m)],[=-sum_(bckm)epsi_(cba)epsi_(ckm)(del)/(delx^(b))*((x^(k))/(r^(3)))J^(m)],[(E.427)=-sum_(bckm)epsi_(cba)epsi_(ckm)(r^(3)(delx^(k))/(delx^(b))-3x^(k)x^(b)r)(J^(m))/(r^(6)).]:}\begin{align*}
(\vec{\nabla} \times \zeta)_{a} & =\sum_{b c} \varepsilon_{a b c} \nabla_{b} \zeta_{c} \\
& =\sum_{b c k m} \varepsilon_{a b c} \frac{\partial}{\partial x^{b}} \frac{1}{r^{3}} \varepsilon_{c k m} x^{k} J^{m} \\
& =-\sum_{b c k m} \varepsilon_{c b a} \varepsilon_{c k m} \frac{\partial}{\partial x^{b}} \cdot\left(\frac{x^{k}}{r^{3}}\right) J^{m} \\
& =-\sum_{b c k m} \varepsilon_{c b a} \varepsilon_{c k m}\left(r^{3} \frac{\partial x^{k}}{\partial x^{b}}-3 x^{k} x^{b} r\right) \frac{J^{m}}{r^{6}} . \tag{E.427}
\end{align*}
Then use the rule that sum_(c)epsi_(cba)epsi_(ckm)=delta_(k)^(b)delta_(m)^(a)-delta_(m)^(b)delta_(k)^(a)\sum_{c} \varepsilon_{c b a} \varepsilon_{c k m}=\delta_{k}^{b} \delta_{m}^{a}-\delta_{m}^{b} \delta_{k}^{a} to (47.4) (a) The centre of mass is defined by m_(1)a_(1)=m_(2)a_(2)m_{1} a_{1}=m_{2} a_{2} obtain
from which the dipolar expression follows on doing the sum over bb and translating back into vector notation.
(46.2) (c) For example, term 5 can be written as
The first term on the right is a divergence which vanishes on the boundary. The second term is zero because del_(alpha)h^(alpha beta)=0\partial_{\alpha} h^{\alpha \beta}=0. The other terms vanish through similar arguments.
(46.3) (c) The first term on the right-hand side of eqn 46.89 is a total derivative and we must have that T^(mu nu)T^{\mu \nu} and its time derivative vanish at the boundary of the mass.
(46.4) (b) Taking the derivative of part (a), we have
(d) For (i) take the solar-system mass to be the solar mass, and the size to be 60 astronomical units (A.U.) where one A.U. ~~1.5 xx10^(11)m\approx 1.5 \times 10^{11} \mathrm{~m}. (This is about twice the radius of Neptune's orbit.). Plugging in yields ~~5kW\approx 5 \mathrm{~kW}. For (ii), if we take the size of the system to be ~~4r_(S)\approx 4 r_{\mathrm{S}}, then we obtain ~~10^(49)W\approx 10^{49} \mathrm{~W}. For (iii), taking M=0.5kgM=0.5 \mathrm{~kg} and R=0.2mR=0.2 \mathrm{~m}, yields ~~10^(-81)W\approx 10^{-81} \mathrm{~W}.
(46.5) (c) TrM_(kℓ)^(TT)=M_(kk)^(TT)=(P_(ik)P_(jk)-(1)/(2)P_(kk)P_(ij))M_(ij)=\operatorname{Tr} M_{k \ell}^{\mathrm{TT}}=M_{k k}^{\mathrm{TT}}=\left(P_{i k} P_{j k}-\frac{1}{2} P_{k k} P_{i j}\right) M_{i j}=(1-(1)/(2)P_(kk))M_(ij)=0\left(1-\frac{1}{2} P_{k k}\right) M_{i j}=0 since P_(kk)=delta_(kk)-n_(k)n_(k)=3-1=2P_{k k}=\delta_{k k}-n_{k} n_{k}=3-1=2. (e) Take n_(1)=sin theta cos phi,n_(2)=sin theta sin phin_{1}=\sin \theta \cos \phi, n_{2}=\sin \theta \sin \phi and n_(3)=n_{3}=cos theta\cos \theta and dOmega=sin thetadthetadphi\mathrm{d} \Omega=\sin \theta \mathrm{d} \theta \mathrm{d} \phi. Then you can show that intn_(1)n_(2)dOmega=0,intn_(3)^(2)dOmega=4pi//3,intn_(1)^(2)n_(2)^(2)dOmega=4pi//15\int n_{1} n_{2} \mathrm{~d} \Omega=0, \int n_{3}^{2} \mathrm{~d} \Omega=4 \pi / 3, \int n_{1}^{2} n_{2}^{2} \mathrm{~d} \Omega=4 \pi / 15, intn_(1)n_(2)n_(3)^(2)dOmega=0\int n_{1} n_{2} n_{3}^{2} \mathrm{~d} \Omega=0, and intn_(3)^(4)dOmega=4pi//5\int n_{3}^{4} \mathrm{~d} \Omega=4 \pi / 5. These are all the possibilities you need since other permutations of the possibilities you need since otry.
indices are related by symmetry
(46.6) (a) Using Newtonian mechanics, the force between the stars is GM^(2)//4r^(2)G M^{2} / 4 r^{2} and equating this to the centripetal force we obtain v^(2)=GM//4rv^{2}=G M / 4 r. Identifying omega=v//r\omega=v / r, the answer follows.
(47.2) The combinations are
corresponding to h=2h=2 and -2 , respectively.
where the separation is a=a_(1)+a_(2)=a_(1)(1+m_(1)//m_(2))a=a_{1}+a_{2}=a_{1}\left(1+m_{1} / m_{2}\right). The force on m_(1)m_{1} obeys m_(1)omega^(2)a_(1)=Gm_(1)m_(2)//a^(2)m_{1} \omega^{2} a_{1}=G m_{1} m_{2} / a^{2}, so omega^(2)=G(m_(1)+m_(2))//a^(3)\omega^{2}=G\left(m_{1}+m_{2}\right) / a^{3}. The moment of inertia II is I=m_(1)a_(1)^(2)+m_(2)a^(2)=m_(1)m_(2)a^(2)//(m_(1)+m_(2))I=m_{1} a_{1}^{2}+m_{2} a^{2}=m_{1} m_{2} a^{2} /\left(m_{1}+m_{2}\right) after some algebra. The gravitational wave luminosity takes the form (32 G//5c^(5))I^(2)omega^(6)\left(32 G / 5 c^{5}\right) I^{2} \omega^{6} and hence the required answer is obtained. Plugging in numbers gives about 200 W of power. The energy of one graviton is ℏomega\hbar \omega so this means about 10^(43)10^{43} gravitons per second are emitted.
(b) Handwaving answer: Very roughly, using a mass ∼1kg\sim 1 \mathrm{~kg}, a length of ∼1m\sim 1 \mathrm{~m}, and a frantic waving frequency of 2 Hz , so that omega∼10s^(-1)\omega \sim 10 \mathrm{~s}^{-1}, yields a luminosity of only of 2 Hz, so that omega∼10s^(-1)\omega \sim 10 \mathrm{~s}^{-1}, yields a luminosity of only ∼10^(-45)W\sim 10^{-45} \mathrm{~W}. The graviton energy is ℏomega∼10^(-33)J\hbar \omega \sim 10^{-33} \mathrm{~J}, so you will need to wave your hands for tens of thousands you will need to wave your hands for tens of
of years before you emit a single graviton.
(48.1) (a) The exterior derivative yields
Note that the derivative with respect to the fifth component vanishes, so we need only sum indices over the usual four dimensions. The expression can be rewritten as
from which the answer follows on identifying dx^( bar(beta))=omega^( hat(beta))\boldsymbol{d} x^{\bar{\beta}}=\boldsymbol{\omega}^{\hat{\beta}}. (b) Using idea 1, we have for the 1 -forms in fivedimensional space
where the sum over the Roman index hat(a)\hat{a} is assumed to range over all five dimensions. From part (a) we have omega^( hat(5))_( hat(alpha))=(1)/(2)F_( hat(alpha) hat(beta))omega^( hat(beta))\boldsymbol{\omega}^{\hat{5}}{ }_{\hat{\alpha}}=\frac{1}{2} F_{\hat{\alpha} \hat{\beta}} \boldsymbol{\omega}^{\hat{\beta}}, from which we can deduce omega_( hat(5))^( hat(alpha))=-(1)/(2)F^( hat(alpha))omega^( hat(beta))\boldsymbol{\omega}_{\hat{5}}^{\hat{\alpha}}=-\frac{1}{2} F^{\hat{\alpha}} \boldsymbol{\omega}^{\hat{\beta}}. We therefore have
In four-dimensional space, we have basis 1 -forms Omega^( hat(alpha))\boldsymbol{\Omega}^{\hat{\alpha}} and so we write -dOmega^( hat(alpha))=Omega^( hat(alpha))_( hat(beta))^^Omega^( hat(beta))-\boldsymbol{d} \boldsymbol{\Omega}^{\hat{\alpha}}=\boldsymbol{\Omega}^{\hat{\alpha}}{ }_{\hat{\beta}} \wedge \boldsymbol{\Omega}^{\hat{\beta}}. However, it is also true that Omega^( hat(alpha))=omega^( hat(alpha))\boldsymbol{\Omega}^{\hat{\alpha}}=\boldsymbol{\omega}^{\hat{\alpha}}, so domega^( hat(alpha))=dOmega^( hat(alpha))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\alpha}}=\boldsymbol{d} \boldsymbol{\Omega}^{\hat{\alpha}}, and we conclude
(49.1) The wavelength lambda\lambda can be estimated using E=hc//lambdaE=h c / \lambda and this yields lambda∼ℓ_(P)\lambda \sim \ell_{\mathrm{P}}, ignoring small numerical constants. The gravitational self-energy for a particle of mass m_(P)m_{P} and 'size' ∼lambda\sim \lambda is ~~Gm_(P)^(2)//lambda\approx G m_{\mathrm{P}}^{2} / \lambda, which also comes out to be ∼E_(P)\sim E_{\mathrm{P}}. By the same argument, this particle would have a Compton wavelength h//m_(P)ch / m_{\mathrm{P}} c and Schwarzschild radius 2Gm_(P)//c^(2)2 G m_{\mathrm{P}} / \mathrm{c}^{2} would also be of the order of ℓ_(P)\ell_{\mathrm{P}}.
(49.2) From eqn 49.17 the action is proportional to the area. Recall that we showed that elements formed by combinations like dtaudsigmasqrt(-gamma)\mathrm{d} \tau \mathrm{d} \sigma \sqrt{-\gamma} are invariants, meaning that we'll obtain the same answer if we use different coordinates. This is just what we mean by reparametrizing the string.
(49.5) (a) The Lagrangian is
If we then use delX^(i)//del sigma=(ds//dsigma)(delX^(i)//del s)\partial X^{i} / \partial \sigma=(\mathrm{d} s / \mathrm{d} \sigma)\left(\partial X^{i} / \partial s\right) and (del vec(X)//del s)^(2)=1(\partial \vec{X} / \partial s)^{2}=1, we have
This has the form of E=m//sqrt(1-v^(2))E=m / \sqrt{1-v^{2}} for transverse motion, with a rest mass given by the string tension, so makes sense if interpreted as an energy.
(49.7) (a) The computation follows, e.g. Example 36.4. We obtain R_( hat(theta) hat(theta))=R_( hat(phi) hat(phi))=1//r_(0)^(2)R_{\hat{\theta} \hat{\theta}}=R_{\hat{\phi} \hat{\phi}}=1 / r_{0}^{2} and R=2//r_(0)^(2)R=2 / r_{0}^{2}. The Einstein equation is solved with 8pi rho=1//r_(0)^(2)8 \pi \rho=1 / r_{0}^{2}.
(b) The cross-sectional area of the string is int_(0)^(theta_(m))r_(0)dthetaint_(0)^(2pi)r_(0)sin thetadphi=2pir_(0)^(2)(1-cos theta_(m))\int_{0}^{\theta_{\mathrm{m}}} r_{0} \mathrm{~d} \theta \int_{0}^{2 \pi} r_{0} \sin \theta \mathrm{~d} \phi=2 \pi r_{0}^{2}\left(1-\cos \theta_{\mathrm{m}}\right). (E.451)
The mass per unit length is then 2pir_(0)^(2)(1-cos theta_(m))rho2 \pi r_{0}^{2}\left(1-\cos \theta_{\mathrm{m}}\right) \rho or, using the previous part, (1-cos theta_(m))//4\left(1-\cos \theta_{\mathrm{m}}\right) / 4.
(49.8) (a) The flat cylindrical metric follows from the substitution
(b) The new variable phi^(')\phi^{\prime} has a range 0 <= phi^(') <= 2pi cos theta_(m)0 \leq \phi^{\prime} \leq 2 \pi \cos \theta_{\mathrm{m}}. As a result, we have intdphi^(')sqrt(g_(phi^(')phi^(')))=2pi a cos theta_(m)\int \mathrm{d} \phi^{\prime} \sqrt{g_{\phi^{\prime} \phi^{\prime}}}=2 \pi a \cos \theta_{\mathrm{m}}. Despite the spacetime looking flat, the fact that this circumference is less than 2pi a2 \pi a gives a sense of the curvature caused by the string.
(49.9) vec(L)_(1)=(1)/(2) vec(a)xx vec(b), vec(L)_(2)=(1)/(2) vec(b)xx vec(c), vec(L)_(3)=(1)/(2) vec(c)xx vec(a)\vec{L}_{1}=\frac{1}{2} \vec{a} \times \vec{b}, \vec{L}_{2}=\frac{1}{2} \vec{b} \times \vec{c}, \vec{L}_{3}=\frac{1}{2} \vec{c} \times \vec{a}, and vec(L)_(4)=(1)/(2)( vec(b)- vec(c))xx( vec(a)- vec(c))=(1)/(2)( vec(b)xx vec(a)- vec(c)xx vec(a)- vec(b)xx vec(c))\vec{L}_{4}=\frac{1}{2}(\vec{b}-\vec{c}) \times(\vec{a}-\vec{c})=\frac{1}{2}(\vec{b} \times \vec{a}-\vec{c} \times \vec{a}-\vec{b} \times \vec{c}), and so the closure property is trivially satisfied.
(49.10) A tetrahedron is one sixth of the volume of the parallelepiped generated by vec(a)xx vec(b)* vec(c)\vec{a} \times \vec{b} \cdot \vec{c}. The modulus sign is because the scalar triple product can be negative debecause the scalar triple product can be negative de-
pending on how the vectors are oriented, but volume pending on how the vectors are oriented, but volume
can only be positive. Substitution of the results from can only be positive. Sub
the previous problem give
the result follows.
(49.12) Substitute the given coordinates into the line element ds^(2)=-dT^(2)+dX^(2)+dY^(2)-dW^(2)\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}+\mathrm{d} Y^{2}-\mathrm{d} W^{2}. The answer follows after several lines of algebra.
(50.1) (a) P_(mu nu)=g_(mu nu)+u_(mu)u_(nu)P_{\mu \nu}=g_{\mu \nu}+u_{\mu} u_{\nu}.
(b) In components, we have P_(mu nu)v^(nu)=v_(mu)+u_(mu)u_(nu)v^(nu)P_{\mu \nu} v^{\nu}=v_{\mu}+u_{\mu} u_{\nu} v^{\nu}. Then
since u_(mu)u^(mu)=-1u_{\mu} u^{\mu}=-1.
(c) P^(mu nu)P_(mu nu)=g^(mu nu)(g_(mu nu)u_(mu)u_(nu))+u^(mu)u^(nu)(g_(mu nu)u_(mu)u_(nu))=4-P^{\mu \nu} P_{\mu \nu}=g^{\mu \nu}\left(g_{\mu \nu} u_{\mu} u_{\nu}\right)+u^{\mu} u^{\nu}\left(g_{\mu \nu} u_{\mu} u_{\nu}\right)=4-1-1+1=31-1+1=3.
(d) P^(alpha beta)u_(alpha;beta)=g^(alpha beta)u_(alpha;beta)+u^(alpha)u^(beta)u_(alpha;beta)P^{\alpha \beta} u_{\alpha ; \beta}=g^{\alpha \beta} u_{\alpha ; \beta}+u^{\alpha} u^{\beta} u_{\alpha ; \beta} for a geodesic. The second term is (grad_(u)u)*u=0\left(\nabla_{\boldsymbol{u}} \boldsymbol{u}\right) \cdot \boldsymbol{u}=0. The first term is u^(beta)_(;beta)=grad*uu^{\beta}{ }_{; \beta}=\boldsymbol{\nabla} \cdot \boldsymbol{u}.
(e) We find
if n*n=1\boldsymbol{n} \cdot \boldsymbol{n}=1.
(50.2) (a) The vector W\boldsymbol{W} which links geodesics is Lie dragged, so we have £_(u)W=[u,W]=0£_{\boldsymbol{u}} \boldsymbol{W}=[\boldsymbol{u}, \boldsymbol{W}]=0, and so
(b) We write u^(mu)u_(mu;nu)=(1)/(2)(u^(mu)u_(mu))_(;nu)=0u^{\mu} u_{\mu ; \nu}=\frac{1}{2}\left(u^{\mu} u_{\mu}\right)_{; \nu}=0, since u^(mu)u_(mu)=u^{\mu} u_{\mu}= -1. The other expression u_(mu;nu)u^(nu)=0u_{\mu ; \nu} u^{\nu}=0 follows from the fact that u\boldsymbol{u} is tangent to a geodesic.
(c) We have
(d) This follows from the symmetry properties of the terms and the results of the previous problem.
(e) Use the result from (d) for the first three terms. For the final term, note R_(beta mu alpha nu)=R_(mu beta nu alpha)R_{\beta \mu \alpha \nu}=R_{\mu \beta \nu \alpha}. The contraction with g^(mu nu)g^{\mu \nu} then turns the Riemann tensor into the Ricci tensor.
(50.3) As in the text for the chapter, we substitute R_(mu nu)u^(mu)u^(nu)=R_{\mu \nu} u^{\mu} u^{\nu}=8pi(T_(mu nu)-(1)/(2)g_(mu nu)T)u^(mu)u^(nu)8 \pi\left(T_{\mu \nu}-\frac{1}{2} g_{\mu \nu} T\right) u^{\mu} u^{\nu}. Considering the strong energy condition then guarantees that the right-hand side of the equation is negative, and so the congruence is forced to contract under the influence of gravity.
(D.1) We can embed a parabolic 2-surface in R^(3)\mathbb{R}^{3} using method II. The parabolic surface in Euclidean 3-space is described by
(D.2) The equation of the surface in R^(4)\mathbb{R}^{4} is given by X^(2)+Y^(2)+X^{2}+Y^{2}+Z^(2)+W^(2)=1Z^{2}+W^{2}=1. If we use spherical coordinates for XX, YY and ZZ we have X=r sin theta cos phi,Y=r sin theta sin phiX=r \sin \theta \cos \phi, Y=r \sin \theta \sin \phi, Z=r cos thetaZ=r \cos \theta and we find that W^(2)=1-r^(2)W^{2}=1-r^{2} and WdW=-rdrW \mathrm{~d} W=-r \mathrm{~d} r. We can use this to eliminate WW from the line element
or
with dOmega_(2)^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega_{2}^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}. This looks very much like the D=2D=2 version, with the replacement of dphi\mathrm{d} \phi with dOmega_(2)\mathrm{d} \Omega_{2}. If we now write r=sin psir=\sin \psi we obtain
(D.3) The equation for the surface may be expressed through the parametric equations
{:[X=(c+a cos v)cos u","],[Y=(c+a cos v)sin u","],[(E.472)Z=a sin v.]:}\begin{align*}
& X=(c+a \cos v) \cos u, \\
& Y=(c+a \cos v) \sin u, \\
& Z=a \sin v . \tag{E.472}
\end{align*}
We can use the method-I rules to determine the induced metric
{:[(del X)/(del u)=-(c+a cos v)sin u",",(del X)/(del v)=-a sin v cos u","],[(del Y)/(del u)=(c+a cos v)cos u",",(del Y)/(del Y)=-a sin v sin u","],[(del Z)/(del u)=0",",(del Z)/(del v)=a cos v","]:}\begin{array}{ll}
\frac{\partial X}{\partial u}=-(c+a \cos v) \sin u, & \frac{\partial X}{\partial v}=-a \sin v \cos u, \\
\frac{\partial Y}{\partial u}=(c+a \cos v) \cos u, & \frac{\partial Y}{\partial Y}=-a \sin v \sin u, \\
\frac{\partial Z}{\partial u}=0, & \frac{\partial Z}{\partial v}=a \cos v,
\end{array}
so